3A 3B 3C 3D 3E 3F Cubic and 3G 3H 3I quartic 3J 3K ...mathsbooks.net/JACPlus Books/11 Methods/Ch03...

109Chapter 3 Cubic and quartic functions

areas oF sTUdy

Use of the notation • y = ƒ(x) for describing the rule of a function and evaluation of ƒ(a), where a is a real number or a symbolic expressionGraphs of polynomial functions to degree 4• Qualitative interpretation of features of graphs, • and families of graphsUse of symbolic notation to develop algebraic • expressions and represent functions, relations and equationsSubstitution into and manipulation, expansion • and factorisation of algebraic expressions, including the remainder and factor theoremsRecognition of equivalent expressions and • simplification of algebraic expressions involving functions and relations, including use of exponent laws and logarithm laws

Determination of rules of simple functions and • relations from given information, including polynomial functions to degree 4Solution of polynomial equations to degree 4, • analytically, numerically and graphicallyThe connection between factors of ƒ(• x), solutions of the equation ƒ(x) = 0 and the horizontal axis intercepts of the graph of the function ƒUse of parameters to represent a family of • functions and general solutions of equations involving these functionsDevelopment of polynomial models, for • example by the use of finite difference tables or solution of a system of simultaneous linear equations obtained from values of a function, or a simple combination of values of a function

eBookpluseBookplus

Digital doc10 Quick Questions

Cubic and quartic functions

3 3A Expanding 3B Long division of polynomials 3C Polynomial values 3D The remainder and factor theorems 3E Factorising polynomials 3F Sum and difference of two cubes 3G Solving polynomial equations 3H Cubic graphs — intercepts method 3I Quartic graphs — intercepts method 3J Graphs of cubic functions in power

function form 3K Domain, range, maximums and

minimums 3L Modelling using technology 3M Finite differences

This chapter will deal mainly with polynomials of degree 3 (cubics). The general equation of a cubic polynomial is P(x) = a3x3 + a2x2 + a1x + a0 or more commonly written as y = ax3 + bx2 + cx + d.

Degree 4 polynomials (quartics) will also be considered. The general equation of a quartic polynomial is P(x) = a4 x4 + a3 x3 + a2 x2 + a1 x + a0 or more commonly written as y = ax4 + bx3 + cx2 + dx + e.

expandingIf we expand three linear factors, for example, (x + 1)(x + 2)(x - 7), we get a cubic polynomial (a polynomial of degree 3) as the following worked example shows.

3a

110

Worked example 1

Expand:a x(x + 2)(x - 3) b (x - 1)(x + 5)(x + 2).

Think WriTe

a 1 Write the expression. a x(x + 2)(x - 3)= x(x2 - 3x + 2x - 6)= x(x2 - x - 6)= x3 - x2 - 6x

2 Expand two linear factors and simplify.

3 Multiply by the remaining factor.

b 1 Write the expression. b (x - 1)(x + 5)(x + 2)= (x - 1)(x2 + 2x + 5x + 10)= (x - 1)(x2 + 7x + 10)= x3 + 7x2 + 10x - x2 - 7x - 10= x3 + 6x2 + 3x - 10

2 Expand two linear factors and simplify.

3 Multiply by the remaining factor and simplify.

Note: Just as there is a shortcut for expanding perfect squares, there is also a shortcut for expanding cubes. We can find the shortcut by expanding (a + b)3 as usual.(a + b)3 = (a + b)(a + b)(a + b) = (a + b)(a2 + 2ab + b2)

= a3 + 2a2b + ab2 + a2b + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3

Similarly, (a - b)3 = a3 - 3a2b + 3ab2 - b3.

Worked example 2

Expand the perfect cube (x - 4)3 using the appropriate rule.

Think WriTe

1 Use the rule (a - b)3 = a3 - 3a2b + 3ab2 + b3. In this case a is x and b is 4.

(x - 4)3

= x3 - 3 × x2 × 4 + 3 × x × 42 - 43

2 Simplify. (x - 4)3

= x3 - 12x2 + 48x - 64

Worked example 3

Expand -2x(x + 5)(x - 12).

Think WriTe/display

1

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a - b)3 = a3 - 3a2b + 3ab2 - b3

maths Quest 11 mathematical methods Cas for the Casio Classpad

On the Main screen, tap:• Action• Transformation• expandComplete the entry line as:expand(-2x(x + 5)(x - 12))Then press E.


2 Write the answer. Expanding -2x(x + 5)(x - 12)= -2x3 + 14x2 + 120x

When expanding three linear factors:expand two factors first, then multiply by the remaining linear factor1. collect like terms at each stage2. (3. x + 2)3 may be written as (x + 2)(x + 2)(x + 2).(4. a ± b)3 = a3 ± 3a2b + 3ab2 ± b3.

rememBer

expanding 1 We 1a Expand each of the following.

a x(x + 6)(x + 1) b x(x - 9)(x + 2) c x(x - 3)(x + 11)d 2x(x + 2)(x + 3) e -3x(x - 4)(x + 4) f 5x(x + 8)(x + 2)g x2(x + 4) h -2x2(7 - x) i (5x)(-6x)(x + 9)j -7x(x + 4)2

2 We 1b Expand each of the following.

a (x + 7)(x + 2)(x + 3) b (x - 2)(x + 4)(x - 5) c (x - 1)(x - 4)(x + 8)d (x - 1)(x - 2)(x - 3) e (x + 6)(x - 1)(x + 1) f (x - 7)(x + 7)(x + 5)g (x + 11)(x + 5)(x - 12) h (x + 5)(x - 1)2 i (x + 2)(x - 7)2

j (x + 1)(x - 1)(x + 1)

3 Expand each of the following.

a (x - 2)(x + 7)(x + 8) b (x + 5)(3x - 1)(x + 4) c (4x - 1)(x + 3)(x - 3)d (5x + 3)(2x - 3)(x - 4) e (1 - 6x)(x + 7)(x + 5) f 3x(7x - 4)(x - 4)g -9x(1 - 2x)(3x + 8) h (6x + 5)(2x - 7)2 i (3 - 4x)(2 - x)(5x + 9)j 2(7 + 2x)(x + 3)(x + 4)

4 We2 Expand using the appropriate rule for expanding cubes.

a (x + 2)3 b (x + 5)3

c (x - 1)3 d (x - 3)3

e (2x - 6)3 f (3x + 4)3

5 We3 Expand using a CAS calculator.a (x + 5)(x - 11)(x + 2) b 3x(x + 6)(x - 1)c 6(x - 5)(-x + 15)(x + 8) d (-x + 5)(x - 12)2

e -x(x + 10)2 f -(x + 13)3

long division of polynomialsThe reverse of expanding is factorising (expressing a polynomial as a product of its linear factors). Before learning how to factorise cubics, you must be familiar with long division of polynomials. You may remember in earlier levels doing long division questions.

exerCise

3a

3B

112

Consider 745 ÷ 3, or )3 745

The process used is as follows.

3 into 7 goes 2 times. Write 2 at the top.2 × 3 = 6. Write down the 6.Subtract to get 1.

Bring down the 4 to form 14.

3 into 14 goes 4 times. Write 4 at the top. 4 × 3 = 12. Write down the 12.Subtract to get 2.

Bring down the 5 to form 25.

3 into 25 goes 8 times. Write 8 at the top.8 × 3 = 24. Write down the 24.Subtract to get 1.Answer: 745 ÷ 3 = 248 remainder 1

The same process can be used to divide polynomials by polynomial factors.

Consider (x3 + 2x2 - 13x + 10) ÷ (x - 3) or )x x x x− + − +3 2 13 103 2

x into x3 goes x2 times x2

)x x x x− +x x− +x x − +x x− +x x3 2)3 2)x x3 2x x)x x)3 2)x x)− +3 2− +)− +)3 2)− +)x x− +x x3 2x x− +x x)x x)− +)x x)3 2)x x)− +)x x) 13x x13x x− +13− +x x− +x x13x x− +x x 103 23 23 23 2− +3 2− +3 2− +3 2− + - (x3 - 3x2) 5x2 - 13x

(consider only the leading terms). Write x2 at the top.x2 × (x - 3) = x3 - 3x2

Write down the x3 - 3x2.

Subtract.(x3 - x3 = 0, 2x2 - -3x2 = 5x2)

Bring down the -13x.

x into 5x2 goes 5x times. Write + 5x at the top. x2 + 5x + 2

)x x x x− + − +3 2 13 103 2

- (x3 - 3x2) 5x2 - 13x - (5x2 - 15x) 2x + 10 - (2x - 6) 16

Quotient

Remainder

5x × (x - 3) = 5x2 - 15x

Write down the 5x2 - 15x.Subtract.Note: 5x2 - 5x2 = 0, -13x - -15x = +2x

Bring down the 10.

x into 2x goes 2 times. Write + 2 at the top.2 × (x - 3) = 2x - 6

Write down the 2x - 6.Subtract to get 16.

Answer: (x3 + 2x2 - 13x + 10) ÷ (x - 3) = x2 + 5x + 2 remainder 16

)3 7452

- 614

)3 745248

- 614

- 1225

- 241

QuotientDividendDividend

Remainder

3Divisor



Worked example 4

Perform the following long divisions and state the quotient and remainder.a (2x3 + 6x2 - 3x + 2) ÷ (x - 6)b (x3 - 7x + 1) ÷ (x + 5)

Think WriTe

a 1 Write the question in long division format. a 2x2x2 2 + 18x + 105x + 105xx - 6)2x2x2 3 + 6x2 - 3x + 2 - (2

) (2)

x (2x (2 3 - 12x12x12 2) 18x2 - 3x - (18x2 - 108x) 105x + 2 - (105x - 630) 632 R

Q2 Perform the long division process.

3 Write down the quotient and remainder. Quotient is 2x2 + 18x + 105, remainder is 632.

b 1 Write the question in long division format. Note that there is no x2 term in this equation. Include 0x2 as a ‘place holder’.

b x2 - 5x + 18x + 5)x3 + 0x2 - 7x + 1 - (x3 + 5x2) -5x2 - 7x - (-5x2 - 25x) 18x + 1 - (18x + 90) -89 R

Q

2 Perform the long division process.

3 Write down the quotient and remainder. Quotient is x2 - 5x + 18, remainder is -89.

Worked example 5

Find the quotient and remainder when x4 - 3x3 + 2x2 - 8 is divided by x + 2.

Think WriTe

1 Write the question in long division format. Include 0x as a ‘place holder’.

x3 - 5x2 + 12x - 24x + 2)x4 - 3x3 + 2x2 + 0x - 8 - (x4 + 2x3) -5x3 + 2x2

- (-5x3 - 10x2) 12x2 + 0x - (12x2 + 24x) -24x - 8 - (-24x - 48) 40The quotient is x3 - 5x2 + 12x - 24.The remainder is 40.

2 Divide x into x4 and write the result above.

3 Multiply x3 by x + 2 and write the result underneath.

4 Subtract and then bring down the next term.

5 Continue to perform the long division process (as you did for cubic polynomials).

6 Write down the quotient and remainder.

eBookpluseBookplus

Tutorialint-0281

Worked example 4

114

Long division of polynomials can be performed on a CAS calculator.

Worked example 6

Calculate the quotient and remainder when x3 - 4x2 - 7x - 5 is divided by x - 1.

Think WriTe/display

1

2 Write the answer given by the calculator. (x3 - 4x2 - 7x - 5) ÷ ( x - 1) = −

−15

1x + x2 - 3x - 10

3 Write the answer required. The quotient is x2 - 3x - 10; the remainder is -15.

Long division of polynomials is similar to long division with numbers.The highest power term is the main one considered at each stage.Key steps are:

How many?1. Multiply and write the result underneath.2. Subtract.3. Bring down the next term.4. Repeat the process until no pronumerals remain to be divided.5. State the quotient and remainder.6.

rememBer

long division of polynomials 1 We4a Perform the following long divisions, and state the quotient and remainder.

a (x3 + 6x2 + 3x + 1) ÷ (x + 3) b (x3 + 4x2 + 3x + 4) ÷ (x + 2)c (x3 + x2 + x + 3) ÷ (x + 1) d (x3 + x2 + 4x + 1) ÷ (x + 2)e (x3 + 2x2 - 5x - 9) ÷ (x - 2) f (x3 + x2 - 9x - 5) ÷ (x - 2)g (x3 - 5x2 + 3x - 8) ÷ (x - 3) h (x3 - 9x2 + 2x - 1) ÷ (x - 5)i 3x3 - x2 + 6x + 5, x + 2 j 4x3 - 4x2 + 10x - 4, x + 1k 2x3 - 7x2 + 9x + 1, x - 2 l 2x3 + 8x2 - 9x - 1, x + 4

exerCise

3B


On the Main screen, tap:• Action• Transformation• propFracComplete the entry line as:propFrac((x3 - 4x2 - 7x - 5)/(x - 1))Then press E.


2 Divide the first polynomial by the second, and state the quotient and remainder.

a 6x3 - 7x2 + 4x + 4, 2x - 1 b 6x3 + 23x2 + 2x - 31, 3x + 4c 8x3 + 6x2 - 39x - 13, 2x + 5 d 2x3 - 15x2 + 34x - 13, 2x - 7e 3x3 + 5x2 - 16x - 23, 3x + 2 f 9x3 - 6x2 - 5x + 9, 3x - 4

3 State the quotient and remainder for each of the following.

a − − − −

+x x x

x

3 26 7 161

b − + + −

−3 7 10 15

3

3 2x x xx

c − + + +

+2 9 17 15

2 1

3 2x x xx

d 4 20 23 2

2 3

3 2x x x

x

− + −+−

4 We4b State the quotient and remainder for each of the following.

a (x3 - 3x + 1) ÷ (x + 1) b (x3 + 2x2 - 7) ÷ (x + 2)c (x3 - 5x2 + 2x) ÷ (x - 4) d (-x3 - 7x + 8) ÷ (x - 1)e (5x2 + 13x + 1) ÷ (x + 3) f (2x3 + 8x2 - 4) ÷ (x + 5)g (-2x3 - x + 2) ÷ (x - 2) h (-4x3 + 6x2 + 2x) ÷ (2x + 1)

5 We5 Find the quotient and remainder for each of the following.

a (x4 + x3 + 3x2 – 7x) ÷ (x – 1) b (x4 – 13x2 + 36) ÷ (x – 2)c (6x4 – x3 + 2x2 – 4x) ÷ (x – 3)

6 We6 Calculate the quotient and remainder for each of the following.a (x3 + 9x2 + 11x + 25) ÷ (x + 15) b (2x3 - 18x2 + 5x - 9) ÷ (x - 31)c (12x3 + 32x - 9) ÷ (3x + 4) d (18x4 + 3x3 + 45) ÷ (2x + 7)

polynomial valuesConsider the polynomial P(x) = x3 - 5x2 + x + 1.

The value of the polynomial when x = 3 is denoted by P(3) and is found by substi tuting x = 3 into the equation in place of x.

That is, P(3) = (3)3 - 5(3)2 + (3) + 1 = 27 - 5(9) + 3 + 1 = 27 - 45 + 4 = -14.

Worked example 7

If P(x) = 2x3 + x2 - 3x - 4, determine:a P(1) b P(-2) c P(a) d P(x + 1).

Think WriTe

a 1 Write the expression. a P(x) = 2x3 + x2 - 3x - 4 P(1) = 2(1)3 + (1)2 - 3(1) - 4 = 2 + 1 - 3 - 4 = -4

2 Replace x with 1.

3 Simplify.

b 1 Write the expression. b P(x) = 2x3 + x2 - 3x - 4 P(-2) = 2(-2)3 + (-2)2 - 3(-2) - 4 = 2(-8) + (4) + 6 - 4 = -16 + 4 + 6 - 4 = -10

2 Replace x with -2.

3 Simplify.

3C

116

c 1 Write the expression. c P(x) = 2x3 + x2 - 3x - 4P(a) = 2a3 + a2 - 3a - 4 2 Replace x with a.

No further simplification is possible.

d 1

2

3

Write the expression.

Replace x with (x + 1).

Expand the right-hand side and collect like terms. Use the rules for expanding cubics and quadratics.

d P(x) = 2x3 + x2 - 3x - 4 P(x + 1) = 2(x + 1)3 + (x + 1)2 - 3(x + 1) - 4

= 2(x3 + 3x2 + 3x + 1) + x2 + 2x + 1 - 3x - 3 - 4 = 2x3 + 6x2 + 6x + 2 + x2 - x - 6 = 2x3 + 7x2 + 5x - 4

Using a CAS calculator, it is best to define the polynomial first. Then finding the value of the polynomial for different values of x can be done very efficiently.

Worked example 8

Use a CAS calculator to determine the following, considering P(x) = 16x4 + 3x3 - 22x + 17.a P(-14) b P(y + 7)

Think WriTe/display

a 1 a

2 Write the answer. P(-14) = 606 749

b 1 b

2 Write the answer. P(y + 7) = 16y4 + 451y3 + 4767y2 + 22 371y + 39 308


On the Main screen, tap:• Action• Command• DefineComplete the entry line as:Define p(x) = 16x4 + 3x3 – 22x + 17P(-14)Then press E after each entry.

Complete the entry line as:p(y + 7)Then press E.


P(a) means the value of P(x) when x is replaced by a and the polynomial is evaluated.

rememBer

polynomial values 1 We7 If P(x) = 2x3 - 3x2 + 2x + 10, determine the following.

a P(0) b P(1) c P(2)d P(3) e P(-1) f P(-2) g P(-3) h P(a) i P(2b)j P(x + 2) k P(x - 3) l P(-4y)

2 We8 Use a CAS calculator to determine the following considering P(x) = 16x4 + 3x3 - 22x + 17.a P(-11) b P(102)c P(2x + 9) d P(x3 + 2)

3 Copy the following table.

Column 1

Column 2

Column 3

Column 4

Column 5

Column 6

Column 7

Column 8

Column 9

P(x) P(1) P(2) P(-1) P(-2)

Rem. when

dividedby

(x - 1)

Rem. when

dividedby

(x - 2)

Rem. when

divided by

(x + 1)

Rem. when

divided by

(x + 2)

a

b

c

d

Complete columns 2 to 5 of the table for each of the following polynomials.a P(x) = x3 + x2 + x + 1 b P(x) = x3 + 2x2 + 5x + 2c P(x) = x3 - x2 + 4x - 1 d P(x) = x3 - 4x2 - 7x + 3

4 Find the remainder when each polynomial in question 2 is divided by (x - 1) and complete column 6 of the table.

5 Find the remainder when each polynomial in question 2 is divided by (x - 2) and complete column 7 of the table.

6 Find the remainder when each polynomial in question 2 is divided by (x + 1) and complete column 8 of the table.

7 Find the remainder when each polynomial in question 2 is divided by (x + 2) and complete column 9 of the table.

8 Copy and complete using your answers to questions 3 to 7 to find the pattern.

a A quick way of fi nding the remainder when P(x) is divided by (x + 8) is to calculate .

b A quick way of fi nding the remainder when P(x) is divided by (x - 7) is to calculate .

c A quick way of fi nding the remainder when P(x) is divided by (x - a) is to calculate .

exerCise

3C

eBookpluseBookplus

Digital docSpreadsheet 016Cubic valuer

eBookpluseBookplus

Digital docWorkSHEET 3.1

118

The remainder and factor theoremsThe remainder theoremIn the previous exercise, you may have noticed that:

The remainder when P(x) is divided by (x - a) is equal to P(a).

That is, R = P(a).This is called ‘the remainder theorem’.We could have derived this result as follows.If 13 is divided by 4, the quotient is 3, and the remainder is 1. That is,

13 ÷ 4 = 3 + 14 and

13 = 4 × 3 + 1.

Similarly, if P(x) = x3 + x2 + x + 1 is divided by (x - 2), the quotient is x2 + 3x + 7 and the remainder is 15. That is,

(x3 + x2 + x + 1) ÷ (x - 2) = x2 + 3x + 7 + 15

2x − and

(x3 + x2 + x + 1) = (x2 + 3x + 7)(x - 2) + 15.

In general, if P(x) is divided by (x - a), and the quotient is Q(x), and the remainder is R, we can write

P(x) ÷ (x - a) = Q(x) + Rx a( )−

and

P(x) = (x - a)Q(x) + R.

Substituting x = a into this last expression yields

P(a) = (a - a)Q(x) + R = 0 × Q(x) + R = R as before.

Worked example 9

Without actually dividing, find the remainder when x3 - 7x2 - 2x + 4 is divided by:a x - 3 b x + 6.

Think WriTe

a 1 Name the polynomial. a Let P(x) = x3 - 7x2 - 2x + 4

2 The remainder when P(x) is divided by (x - 3) is equal to P(3).

R = P(3) = 33 - 7(3)2 - 2(3) + 4 = 27 - 7(9) - 6 + 4 = 27 - 63 - 6 + 4 = -38

b The remainder when P(x) is divided by (x + 6) is equal to P(-6).

b R = P(-6) = (-6)3 - 7(-6)2 - 2(-6) + 4 = -216 - 7(36) + 12 + 4 = -216 - 252 + 12 + 4 = -452

3d



Worked example 10

The remainder when x3 + kx2 + x - 2 is divided by (x - 2) is equal to 20. Find the value of k.

Think WriTe

1 Name the polynomial. Let P(x) = x3 + kx2 + x - 2.R = P(2)

= 23 + k(2)2 + 2 - 2= 8 + 4k

2 The remainder when P(x) is divided by (x - 2) is equal to P(2).

3 We are given R = 20. Put 8 + 4k = 20.

Since R = 20. 8 + 4k = 20 4k = 12 k = 3

4 Solve for k.

The factor theoremThe remainder when 12 is divided by 4 is zero, since 4 is a factor of 12.

Similarly, if the remainder (R) when P(x) is divided by (x - a) is zero, then (x - a) must be a factor of P(x).

Since R = P(a), all we need to do is to find a value of a that makes P(a) = 0, and we can say that (x - a) is a factor.

If P(a) = 0, then (x - a) is a factor of P(x).

This is called ‘the factor theorem’.Imagine P(x) could be factorised as follows:

P(x) = (x - a)Q(x), where Q(x) is ‘the other’ factor of P(x).Then we have P(a) = (a - a)Q(a) = 0 × Q(a) = 0.

So if P(a) = 0, (x - a) is a factor.

Worked example 11

Apply the factor theorem to determine which of the following is a factor of x4 - 4x3 - 43x2 + 58x + 240.a (x + 2) b (x - 1)

Think WriTe

a 1 Name the polynomial. a Let P(x) = x4 - 4x3 - 43x2 + 58x + 240

2 To find the remainder when P(x) is divided by (x - a), find P(a).

P(-2) = (-2)4 - 4(-2)3 - 43(-2)2 + 58(-2) + 240 = 16 - 4(-8) - 43(4) - 116 + 240 = 16 + 32 - 172 - 116 + 240 = 0

3 State the answer. As P(-2) = 0, the remainder when P(x) is divided by (x + 2) is zero so (x + 2) is a factor.

b 1 To find the remainder when P(x) is divided by (x - a), find P(a).

b P(1) = (1)4 - 4(1)3 - 43(1)2 + 58(1) + 240 = 1 - 4 - 43 + 58 + 240 = 252

2 State the answer. As P(1) = 252, the remainder when P(x) is divided by (x - 1) is 252; therefore, (x - 1) is not a factor.

120

Remainder 1. R = P(a) when P(x) is divided by x – a.If 2. P(a) = 0, then (x - a) is a factor of P(x).

rememBer

The remainder and factor theorems 1 We9 Without actually dividing, find the remainder when x3 + 3x2 - 10x - 24 is

divided by:a x - 1 b x + 2 c x - 3 d x + 5e x - 0 f x - k g x + n h x + 3c.

2 Find the remainder when the first polynomial is divided by the second without performing long division.a x3 + 2x2 + 3x + 4, x - 3 b x3 - 4x2 + 2x - 1, x + 1c x3 + 3x2 - 3x + 1, x + 2 d x3 - x2 - 4x - 5, x - 1e 2x3 + 3x2 + 6x + 3, x + 5 f -3x3 - 2x2 + x + 6, x + 1g x3 + x2 + 8, x - 5 h x3 - 3x2 - 2, x - 2i -x3 + 8, x + 3 j x3 + 2x2, x - 7

3 a We 10 The remainder when x3 + kx + 1 is divided by (x + 2) is -19. Find the value of k.b The remainder when x3 + 2x2 + mx + 5 is divided by (x - 2) is 27. Find the value of m.c The remainder when x3 - 3x2 + 2x + n is divided by (x - 1) is 1. Find the value of n.d The remainder when ax3 + 4x2 - 2x + 1 is divided by (x - 3) is -23. Find the value of a.e The remainder when x3 - bx2 - 2x + 1 is divided by (x + 1) is 0. Find the value of b.f The remainder when -4x2 + 2x + 7 is divided by (x - c) is -5. Find a possible whole

number value of c.g The remainder when x2 - 3x + 1 is divided by (x + d) is 11. Find the possible

values of d.h The remainder when x3 + ax2 + bx + 1 is divided by (x - 5) is -14. When the cubic

polynomial is divided by (x + 1), the remainder is -2. Find a and b.

4 We 11 Apply the factor theorem to determine which of the following are factors of x3 + 2x2 - 11x - 12.a (x - 1) b (x - 3) c (x + 1) d (x + 2)

5 Prove that each of the following are linear factors of x3 + 4x2 - 11x - 30 by substi tuting

values into the cubic function: (x + 2), (x - 3), (x + 5).

6 Use the factor theorem to show that the first polynomial is exactly divisible by the second (that is, the second polynomial is a factor of the first).a x3 + 5x2 + 2x - 8, x - 1 b x3 - 7x2 - x + 7, x - 7c x3 - 7x2 + 4x + 12, x - 2 d x3 + 2x2 - 9x - 18, x + 2e x3 + 3x2 - 9x - 27, x + 3 f -x3 + x2 + 9x - 9, x - 1g -2x3 + 9x2 - x - 12, x - 4 h 3x3 + 22x2 + 37x + 10, x + 5

7 mC a When x3 + 2x2 - 5x - 5 is divided by (x + 2), the remainder is:A -5 B -2 C 0 D 2 E 5

b Which of the following is a factor of 2x3 + 15x2 + 22x - 15?A (x - 1) B (x - 2) C (x + 3) D (x - 5) E (x + 4)

c When x3 - 13x2 + 48x - 36 is divided by (x - 1), the remainder is:A -3 B -2 C -1 D 0 E 1

d Which of the following is a factor of x3 - 5x2 - 22x + 56?A (x - 2) B (x + 2) C (x - 5) D (x + 5) E (x + 7)

exerCise

3d

eBookpluseBookplus

Digital docSpreadsheet 016Cubic valuer



8 Find one factor of each of the following cubic polynomials.a x3 - 3x2 + 3x - 1 b x3 - 7x2 + 16x - 12c x3 + x2 - 8x - 12 d x3 + 3x2 - 34x - 120

9 For the polynomial P(x) = 6x3 + 7x2 - x - 2, find:a i P(-1) ii P 1

2( ) iii P −( )23

b i Factorise P(x) as the product of (x + 1) and a quadratic factor. ii Further factorise so P(x) is written as the product of three linear factors.c Explain how the other two linear factors relate to what you found in parts b and c .d Copy and complete the following: In general if (ax + b) is a factor, then P(…) = 0.

Factorising polynomialsUsing long divisionOnce one factor of a polynomial has been found (using the factor theorem as in the previous section), long division may be used to fi nd other factors.

Worked example 12

Use long division to factorise x3 - 19 x + 30.

Think WriTe

1 Name the polynomial.Note: There is no x2 term, so include 0x2.

P(x) = x3 - 19x + 30P(x) = x3 + 0x2 - 19x + 30

2 Look at the last term in P(x), which is 30. This suggests it is worth trying P(5) or P(-5). Try P(-5). P(-5) = 0 so (x + 5) is a factor.

P(-5) = (-5)3 - 19 × (-5) + 30 = -125 + 95 + 30 = 0So (x + 5) is a factor.

3 Divide (x + 5) into P(x) using long division to fi nd a quadratic factor.

x2 - 5x + 6x + 5)x3 + 0x2 - 19x + 30 -(x3 + 5x2) -5x2 - 19x -(-5x2 - 25x) 6x + 30 -(6x + 30) 0

4 Write P(x) as a product of the two factors found so far.

P(x) = (x + 5)(x2 - 5x + 6)

5 Factorise the second bracket if possible. P(x) = (x + 5)(x - 2)(x - 3)

Note: In this example, P(x) may have been factorised without long division by fi nding all three values of x which make P(x) = 0, and hence three factors, then checking that the three factors multiply to give P(x).

Using short divisionThe process of long division can take a lot of time (and space). One short division method is shown here; it may take a little longer to understand, but it is quicker than long division once mastered.

Consider P(x) = x3 + 2x2 - 13x + 10. Using the factor theorem, we can fi nd that (x - 1) is a factor of P(x). So, P(x) = (x - 1)(ax2 + bx + c).

eBookpluseBookplus

Digital docSkillSHEET 3.1

Reviewing the discriminant

3e

eBookpluseBookplus

Tutorialint-0282

Worked example 12

122

Actually, we know more than this: as P(x) begins with x3 and ends with +10, we could write:P(x) = (x - 1)(x2 + bx - 10)

Imagine expanding this version of P(x). Our x2 terms give -1x2 + bx2.Since P(x) = x3 + 2x2 - 13x + 10, we need +2x2. That is, we need -1x2 + 3x2. To get this, the bx

must be 3x, as when x in the first bracket is multiplied by 3x in the second bracket, +3x2 results. That is, we have deduced

P(x) = (x - 1) (x2 + 3x - 10).Factorising the second bracket gives

P(x) = (x - 1)(x + 5)(x - 2)

Worked example 13

Factorise the following using short division where possible.a x3 - 5x2 - 2 x + 24b x4 + x3 - 13 x2 - 25 x - 12

Think WriTe

a 1 Name the polynomial. a Let P(x) = x3 - 5x2 - 2x + 24

2 Look for a value of x such that P(x) = 0. Try P(-2).

P(-2) does equal 0, so (x + 2) is a factor.

P(-2) = (-2)3 - 5 × (-2)2 - 2 × (-2) + 24 = -8 - 20 + 4 + 24 = -28 + 28 = 0So (x + 2) is a factor.

3 Write the original polynomial as the found factor multiplied by ax2 + bx + c. The first term in the brackets must be x2, and the last term must be 12.

P(x) = x3 - 5x2 - 2x + 24 P(x) = (x + 2) (ax2 + bx + c) = (x + 2) (x2 + bx + 12)

4 Imagine the expansion of the expression in step 3. We have 2x2, and require -5x2. We need an extra -7x2. So b = -7.

2x2 + bx2 = - 5x2

b = -7P(x) = (x + 2)(x2 - 7x + 12)

5 Factorise the second bracket if possible. P(x) = (x + 2)(x - 3)(x - 4)

b 1 Name the polynomial. b Let P(x) = x4 + x3 - 13x2 - 25x - 12

2 Look for a value of x such that P(x) = 0.Try P(-1).

P(-1) = 0So (x + 1) is a factor.

3 It is difficult to factorise a quartic using short division so we will use long division here.

x3 + 0x2 - 13x - 12x + 1) x

4 + x3 - 13x2 - 25x - 12 - (x4 + x3) 0 - 13x2 - 25x - (-13x2 + 13x) -12x - 12 - (-12x - 12) 0

4 Name the cubic factor, and try to find another factor using the factor theorem.

Let Q(x) = x3 - 13x - 12.Q(-3) = 0So (x + 3) is a factor.

5 Factorise the cubic using short division. Q(x) = x3 - 13x - 12= (x + 3)(x2 - 3x - 4)



6 Factorise the quadratic if possible. Q(x) = (x + 3)(x - 4)(x + 1)

7 Write the original polynomial in factorised form.

P(x) = x4 + x3 - 13x2 - 25x - 12= (x + 1)(x + 3)(x - 4)(x + 1)= (x + 1)2(x + 3)(x - 4)

Expressions can also be factorised using a CAS calculator.

Worked example 14

Use a CAS calculator to factorise the expression 20 x3 + 4 x2 - 199 x - 210.

Think WriTe/display

1

2 Write the answer. Factorising 20x3 + 4x2 - 199x - 210= (2x - 7)(2x + 5)(5x + 6)

To factorise a polynomial:Let 1. P(x) = the given polynomial.Use the factor theorem to find a linear factor (try factors of the constant term).2. Use long or short division to find another factor.3. Repeat steps 2 and 3, or factorise by inspection if possible. Alternatively, use the factor 4. function on your CAS calculator.

rememBer

Factorising polynomials 1 We 12 Use long division to factorise each dividend.

a )x x x x+ + + +1 10 27 183 2 b )x x x x+ + + +2 8 17 103 2

c )x x x x+ + + +9 12 29 183 2 d )x x x x+ + + +1 8 19 123 2

e )x x x x+ + + +3 14 61 843 2 f )x x x x+ + + +7 12 41 423 2

exerCise

3e

On the Main screen, tap:• Action• Transformation• factorComplete the entry line as:factor(20x3 + 4x2 - 199x - 210)Then press E.

124

g )x x x x+ + + +2 4 5 23 2 h )x x x x+ + + +3 7 16 123 2

i )x x x x+ + + +5 14 65 1003 2 j )x x x x3 213 40+ +

k )x x x x3 27 12+ + l )x x x x+ + +5 10 253 2

m )x x x x+ + +1 6 53 2 n )x x x+ +6 63 2

2 We 12, 13 Factorise the following as fully as possible.

a x3 + x2 - x - 1 b x3 - 2x2 - x + 2c x3 + 7x2 + 11x + 5 d x3 + x2 - 8x - 12e x3 + 9x2 + 24x + 16 f x3 - 5x2 - 4x + 20g x3 + 2x2 - x - 2 h x3 - 7x - 6i x3 + 3x2 - 4 j x3 + x2 + x + 6k x3 + 8x2 + 17x + 10 l x3 + x2 - 9x - 9m x3 - x2 - 8x + 12 n x3 + 9x2 - 12x - 160o x4 + 4x3 + 3x2 - 4x - 4 p x4 + 3x3 - 6x2 - 28x - 24q x4 + 6x3 + 8x2 - 6x - 9 r x4 - 5x3 - 17x2 + 21x

3 We 14 Use a CAS calculator to factorise the following as fully as possible.

a 3x3 - x2 - 10x b 4x3 + 2x2 - 2xc 3x3 - 6x2 - 24x d -2x3 - 12x2 - 18xe 6x3 - 6x2 f -x3 - 7x2 - 12xg -x3 - 3x2 + x + 3 h -2x3 + 10x2 - 12xi -6x3 - 5x2 + 12x - 4 j -5x3 + 24x2 - 36x + 16k -x5 - x4 + 21x3 + 49x2 - 8x - 60 l 24x4 - 53x3 - 71x2 + 152x + 20

4 Factorise the following as fully as possible.

a 2x3 + 5x2 - x - 6 b 3x3 + 14x2 + 7x - 4c 3x3 + 2x2 - 12x - 8 d 4x3 + 35x2 + 84x + 45e 5x3 + 9x2 + 3x - 1 f x3 + x2 + 4g 4x3 + 16x2 + 21x + 9 h 6x3 - 23x2 + 26x - 8i 10x3 + 19x2 - 94x - 40 j 7x3 + 12x2 - 60x + 16k 2x4 - x3 - 11x2 - 11x - 3 l 6x4 + 11x3 - 22x2 - x + 6

sum and difference of two cubesTwo special cases of cubic polynomials, called ‘sum and difference of cubes’ are discussed in this section. There are shortcuts for factorising such cubic expressions. Examples of each are shown in the table below.

Sum of cubes Difference of cubes

x3 + 23 x3 - 27

125 + 64b3 x3

1000 - 81y3

x3y3 + 1 w6 - 1

(2x + 1)3 + 8 216 - (uv)3

eBookpluseBookplus

Digital docSpreadsheet 096Polynomials zero search

3F



Consider the following expansions.(a + b)(a2 - ab + b2) and (a - b)(a2 + ab + b2)

= a3 - a2b + ab2 + ba2 - ab2 + b3 = a3 + a2b + ab2 - ba2 - ab2 - b3

= a3 - a2b + ab2 + a2b - ab2 + b3 = a3 + a2b + ab2 - a2b - ab2 - b3

= a3 + b3 = a3 - b3

These expansions show that:a3 + b3 = (a + b)(a2 - ab + b2) and a3 - b3 = (a - b)(a2 + ab + b2).That is, we have two formulas which may be used to factorise sums and differences of cubes.

Worked example 15

Factorise the following using the sum or difference of cubes formula.

a x3 - 1000y3 b 2(x + 6)3 + 16

Think WriTe

a 1 Write the expression. a x3 - 1000y3

= x3 - (10y)3

a = x, b = 10y

= (x - 10y)[x2 + x(10y) + (10y)2]= (x - 10y)(x2 + 10xy + 100y2)

2 Recognise a difference of cubes.

3 Identify a and b for use with the formula a3 - b3 = (a - b)(a2 + ab + b2).

4 Use the formula to factorise.

5 Simplify.

b 1 Write the expression. b 2(x + 6)3 + 16

= 2[(x + 6)3 + 8]= 2[(x + 6)3 + 23]

a = (x + 6), b = 2

= 2[(x + 6) + 2][(x + 6)2 - (x + 6)(2) + 22]

= 2(x + 8)(x2 + 12x + 36 - 2x - 12 + 4)= 2(x + 8)(x2 + 10x + 28)

2 Take out a common factor of 2 to produce a sum of cubes.

3 Identify a and b for use with the formula a3 + b3 = (a + b)(a2 - ab + b2).

4 Apply the sum of cubes formula.

5 Simplify.

Sum of two cubes: a3 + b3 = (a + b)(a2 - ab + b2)Difference of two cubes: a3 - b3 = (a - b)(a2 + ab + b2)

rememBer

sum and difference of two cubes 1 Identify a and b (as used in the above sum and difference of cubes expressions) in each of the

following (do not factorise).a x3 + 63 b 8y3 + z3 c (x + 5)3 + 27

d 1 - 64h3 e 18 c3 - e3g3 f t3 - u3

216 2 We 15a Factorise the following using the sum or difference of cubes formula.

a x3 - 125 b j3 + k3 c y3 - 8

d 27x3 + y3 e 64t3 - 216u3 f x3 - 1

eBookpluseBookplus

Tutorialint-0283

Worked example 15

exerCise

3F

126

g x3

729 + 8p3 h 27r3 - 1 i (3k)3 - 1

8

j s3t3 + g6

3 We15b Factorise:

a (a - 1)3 + a3 b (x + 2)3 - 8c (2x + 3)3 + 1 d (w - 5)3 - w3

e (2m + p)3 + (3m - p)3 f 27x3 - (x + 3)3

g (2y + 7)3 + (y - 2)3 h (3x + y)3 + (x - 4y)3

i (2 - 4p)3 - (p + 1)3 j (5x - 9)3 - (7 - x)3

k x6 + y9 l 2x3 - 54 m 3a3 + 3 n 6(x2 + 1)3 + 162

4 When mx3 - ny3 is fully factorised it gives (3x − y)(9x2 + 3xy + y2). What are the values of m and n?

5 a Write 3x3 + my3 in the form a3 + b3. b Identify the values of a and b.

c Factorise using the rule for sum of cubes.

solving polynomial equationsisolating xCubic equations of the form a(x - b)3 + c = 0 may be solved by isolating x as follows. a(x - b)3 = -c

(x - b)3 = − c

a

x - b = − c

a3

x = b + − c

a3

Unlike a square root, a cube root can be only positive or negative, but not both, for example,− +83 = -2, 83 = 2.

Worked example 16

Solve 3(x + 2)3 + 192 = 0 by isolating x.

Think WriTe

1 Write the equation. 3(x + 2)3 + 192 = 0

3(x + 2)3 = -192

(x + 2)3 = -64

x + 2 = -4

x = -4 - 2 = -6

2 Subtract 192 from both sides.

3 Divide both sides by 3.

4 Take the cube root of both sides.

5 Subtract 2 from both sides and simplify.

3G



Factorising to solve polynomial equationsThe Null Factor Law applies to cubic and quartic equations just as it does for quadratics.

If P(x) = (x - a)(x - b)(x - c) = 0, then solutions are: x = a, x = b and x = c.If P(x) = k(lx - a)(mx - b)(nx - c) = 0, solutions are found by solving the following

equations:

lx - a = 0, mx - b = 0 and nx - c = 0.

Worked example 17

a x4 = 16x2 b 2x3 - 11x2 + 18x - 9 = 0.

Think WriTe

a 1 Write the equation. a x4 = 16x2

x4 - 16x2 = 0

x2(x2 - 16) = 0

x2(x + 4)(x - 4) = 0

x = 0, x + 4 = 0 or x - 4 = 0so x = 0, x = -4 or x = 4

2 Rearrange so that all terms are on the left.

3 Take out a common factor of x.

4 Factorise the brackets using a difference of squares.

5 Use the Null Factor Law to solve.

b 1 Name the polynomial. b Let P(x) = 2x3 - 11x2 + 18x - 9.P(1) = 2 - 11 + 18 - 9

= 0So (x - 1) is a factor.

2x 2x 2 2 - 9x + 9x + 9xx - 1)2x2x2 3 - 11x2 - 18x - 9

2x2x2 3 - 2x 2x 2 2

-9x2 + 18x-9x2 + 9x

9x - 99x - 9

0P(x) = (x - 1)(2x2 - 9x + 9)P(x) = (x - 1)(2x - 3)(x - 3)

For (x - 1)(2x - 3)(x - 3) = 0

x - 1 = 0, 2x - 3 = 0 or x - 3 = 0

so x = 1, x = 32

or x = 3

2 Use the factor theorem to fi nd a factor (search for a value a such that P(a) = 0). Consider factors of the constant term (that is, factors of 9 such as 1, 3). The simplest value to try is 1.

3 Use long or short division to fi nd the other factors of P(x).

4 Factorise the brackets.

5Consider the factorised equation to solve.

6 Use the Null Factor Law to solve.

Polynomial equations can also be solved using a CAS calculator, which is useful when the solutions are not rational. The solutions can be found by using the solve function on a calculator page, or by constructing a graph.

128

Worked example 18

Use a CAS calculator to solve x3 + 3 x2 - 8 x - 5 = 0.

Think WriTe/display

Method 1: Algebraically

a 1

2 Write the answer. Solving x3 + 3x2 − 8x − 5 = 0 for x gives x = -4.524, -0.536, 2.060.

Method 2: Graphically

b 1

2 Write the answer. Solving x3 + 3x2 − 8x − 5 = 0 for x gives x = -4.524, -0.536, 2.060.

To solve a polynomial equation:Rewrite the equation so it equals zero.1. Factorise the polynomial as much as possible.2. Let each linear factor equal zero and solve for 3. x in each case.

rememBer


On the Main screen, tap:• Action• Advanced• solveComplete the entry line as:solve(x3 + 3x2 - 8x - 5 = 0, x)Then press E.

Construct the graph of y = x3 + 3x2 - 8x - 5 on the Graph & Tab screen. Adjust the window appropriately. We want to find where the curve intersects with y = 0, which is the x-axis. Tap:• Analysis• G-Solve• RootTo find the other roots use the left and right arrow keys.


solving polynomial equations 1 We 16 Solve the following by isolating x.

a 2(x - 1)3 - 250 = 0 b 3(x + 2)3 + 81 = 0c (x - 4)3 - 1000 = 0 d (x + 7)3 - 8 = 0e -2(x - 5)3 - 2 = 0 f -(x + 3)3 + 1 = 0g (2x + 3)3 - 27 = 0 h 4(3x - 1)3 + 500 = 0

i 23 (x + 1)3 + 18 = 0 j 1

2 (5 - x)3 - 32 = 0

k (x - 5)3 = 343 l 4 - 45 (x + 8)3 = 104

2 Find all solutions of the following equations.

a (x - 1)(x - 2)(x - 5) = 0 b (x + 3)(x + 4)(x + 7) = 0c (x - 5)(x + 2)(x - 9) (x - 1) = 0 d (2x - 4)(x + 1)(x - 3) (x + 2) = 0e (3x + 12)(x - 4)(x + 4) = 0 f (2 - x)(x + 2)(1 - x)(1 + x) = 0g (x + 5)(x - 8)2 = 0 h (x - 1)3 = 0i x2 (x + 1)(x - 1) = 0 j -3x(x - 9)3 = 0k (6 - x)2 (2 + x) = 0 l x2(2x + 7) = 0m x(5x - 6)(2x + 3) (6 - 7x) = 0 n (3 - 4x)2 (5x - 1) = 0

3 We 17a Solve.

a x3 - 4x = 0 b x3 - 16x = 0 c 2x4 - 50x2 = 0d -3x4 + 81 = 0 e x3 + 5x2 = 0 f x3 - 2x2 = 0g -4x3 + 8x = 0 h 12x3 + 3x2 = 0 i 4x3 - 20x4 = 0j x4 - 5x3 + 6x2 = 0 k x3 - 8x2 + 16x = 0 l x3 + 6x2 = 7xm 9x2 = 20x + x3 n x3 + 6x = 4x2

4 We 17b Use the Null Factor Law to solve the following.a x3 - x2 - 16x + 16 = 0 b x3 - 6x2 - x + 30 = 0c x3 - x2 - 25x + 25 = 0 d x3 + 4x2 - 4x - 16 = 0e x3 - 4x2 + x + 6 = 0 f x3 - 4x2 - 7x + 10 = 0g x4 - 3x3 - 7x2 + 15x = -18 h x4 + 2x3 - 13x2 + 10xi 2x3 + 15x2 + 19x + 6 = 0 j -4x3 + 16x2 - 9x - 9 = 0k -2x3 - 9x2 - 7x + 6 = 0 l 2x3 + 4x2 - 2x - 4 = 0

5 We 18 Use a CAS calculator to find all solutions to the following equations.a x3 - 17x2 - 56x + 1153 = 0 b x3 + 12x2 - 49x - 588 = 0c -x3 + 17x2 + 65x - 1521 = 0 d x3 – 48x2 + 768x - 4096 = 0e x3 + 6x2 - 6x + 2 = 0 f x3 – 14x2 – 4x + 13 = 0g 3x2 + 2x + 1 = x3 h -2x3 – 3x2 + 2x + 0.5 = 0i x4 + 4x3 - 5x2 + 15 = 0 j 2x4 + 25x3 + 17x = 9

6 mC Which of the following is a solution to x3 - 7x2 + 2x + 40?A -5 B -4 C -2 D 1 E 2

7 mC A solution of x3 - 9x2 + 15x + 25 = 0 is x = 5. How many other (distinct) solutions are there?A 0 B 1 C 2 D 3 E 4

Cubic graphs — intercepts methodA good sketch graph of a function shows:1. x- and y-intercepts,2. the behaviour of the function at extreme values of x, that is, as x approaches infi nity (x → +∞)

and as x approaches negative infi nity (x → -∞), and3. the general location of turning points.

exerCise

3G

eBookpluseBookplus

Digital docWorkSHEET 3.2

3h

130

Note that for cubic functions, ‘humps’ are not symmetrical as they are for parabolas, but are skewed to one side.

The graphs below show the two main types of cubic graph.y

x

Turningpoints

y

x

A positive cubic A negative cubic

For positive cubic graphs, as positive values of x become larger and larger, y-values also become larger.

For negative cubic graphs, as positive values of x become larger and larger, y-values become smaller.

Sometimes instead of two turning points there is a point of inflection, where the graph changes from a decreasing gradient to zero to an increasing gradient (or vice versa). When this occurs, there is only one x-intercept.

Consider the general factorised cubic f (x) = (x - a)(x - b)(x - c).

The Null Factor Law tells us that f (x) = 0 when x = a or x = b or x = c.

The y-intercept occurs when x = 0, that is, the y-intercept is f (0) = (0 - a)(0 - b)(0 - c) = -abc

a

−abc

bc

y

x

Worked example 19

Sketch the following, showing all intercepts:a y = (x - 2)(x - 3)(x + 5) b y = (x - 6)2 (4 - x) c y = (x - 2)3.

Think WriTe/draW

a 1 Note that the function is already factorised and that the graph is a positive cubic.

a y = (x - 2)(x - 3)(x + 5)

Cubic with a point of inflection

Point of inflectiony

x



2 The y-intercept occurs where x = 0. Substitute x = 0 into the equation.

y-intercept: if x = 0,y = (-2)(-3)(5)

= 30Point: (0, 30)

3 Use the Null Factor Law to find the x-intercepts. (Put each bracket = 0 and solve a mini-equation.)

x-intercepts: if y = 0,x - 2 = 0, x - 3 = 0 or x + 5 = 0x = 2, x = 3 or x = -5Points: (2, 0), (3, 0), (-5, 0)

4 Combine information from the above steps to sketch the graph.

−5

30

2 3

y

x

b 1 The graph is a negative cubic (the -x in the last factor produces a negative x3 coefficient if the RHS is expanded).

b y = (x - 6)2(4 - x)

2 Substitute x = 0 to find the y-intercept. y-intercept: if x = 0,y = (-6)2(4)

= 144Point: (0, 144)

3 Use the Null Factor Law to find the x-intercepts. (Put each bracket = 0 and solve a mini-equation.)

x-intercepts: if y = 0, x - 6 = 0 or 4 - x = 0 x = 6 or x = 4Points: (6, 0), (4, 0)

4 Combine all information and sketch the graph. Note the skimming of the x-axis indicative of a repeated factor, in this case the (x - 6)2 part of the expression.

144

4 6

y

x

c 1 Positive cubic. c y = (x - 2)3

2 Substitute x = 0 to find the y-intercept. y-intercept: if x = 0,y = (-2)3

= -8

3 Use the Null Factor Law to find the x-intercept. (Put each bracket = 0 and solve a mini-equation.)

x-intercept: if y = 0, x - 2 = 0 x = 2

132

4 Combine all information and sketch the graph. The cubed factor, (x - 2), indicates a point of infl ection and only one x-intercept.

2

−8

y

x

If a cubic function is not in the form f (x) = (x - a)(x - b)(x - c), we may try to factorise to fi nd the x-intercepts. We can use the factor theorem and division of poly nomials to achieve this.

Worked example 20

Sketch the graph of y = 2x3 - 11x2 + 7x + 20 showing all intercepts.

Think WriTe/draW

1 Write the equation, and name the polynomial, P(x).

y = P(x) = 2x3 - 11x2 + 7x + 20

2 Note the graph is a positive cubic.

3 Let x = 0 to fi nd the y-intercept.Note: All terms involving x are equal to zero.

y-intercept: if x = 0,y = 20Point: (0, 20)

4 Factorise P(x) to fi nd x-intercepts.

(x - 1) is not a factor.

P(1) = 2 × 13 - 11 × 12 + 7 × 1 + 20= 2 - 11 + 7 + 20= 18≠ 0

P(-1) = 2 × (-1)3 - 11 × (-1)2 + 7 × (-1) + 20= -2 - 11 - 7 + 20= 0

So (x + 1) is a factor.

5 Use long or short division to factorise P(x). Here, short division has been used.

P(x) = (x + 1)(2x2 - 13x + 20)= (x + 1)(2x - 5)(x - 4)

6 Write down the x-intercepts (determined by putting each bracket = 0 and solving for x).

x-intercepts: if y = 0,x = -1, 5

2, 4

Points: (-1, 0), (52, 0), (4, 0)

7 Use all available information to sketch the graph.

20

4−1 5–2

y

x

eBookpluseBookplus

Tutorialint-0284

Worked example 20



Worked example 21

Sketch the graph of y = 3x3 + x2 - 2x + 5 using a CAS calculator. Find all intercepts and stationary points.

Think WriTe/display

1

2

3

On the Graph & Tab screen, complete the entry line as:y1 = 3x3 + x2 - 2x + 5Tick the y1 box and tap !.Create an appropriate viewing window; then tap:• Analysis• G-Solve• Root

To find the maximum turning point, tap:• Analysis• G-Solve• Max

To find the minimum turning point, tap:• Analysis• G-Solve• Min

134

4 Sketch the graph showing all relevant points.

(−0.595, 5.912)

(−1.51, 0)

(0.37, 4.55)

0

(0, 5)

y

y = 3x3 + x2 − 2x + 5

x

To sketch a cubic function of the form y = (x + 1)(2x - 5)(x - 4)f (x) = Ax3 + Bx2 + Cx + D:

determine if the expression is a positive or negative 1. cubic (that is, if A is positive or negative)find the 2. y-intercept (let x = 0)factorise if necessary and/or possible, for 3. example, obtain an expression in the form f (x) = (x - a)(x - b)(x - c)find the 4. x-intercepts (let factors of f (x) equal 0)use all available information to sketch the graph.5.

20

4−1 5–2

y

x

rememBer

Cubic graphs — intercepts method 1 We19 Sketch the following, showing all intercepts.

a y = (x - 1)(x - 2)(x - 3) b y = (x + 6)(x + 1)(x - 7)c y = (x + 8)(x - 11)(x + 1) d y = (2x - 5)(x + 4)(x - 3)e y = (4x - 3)(2x + 1)(x - 4) f y = (x - 3)2(x - 6)

2 Sketch the following.a y = (2 - x)(x + 5)(x + 3) b y = (x + 8)(x - 8)(2x + 3)c y = x(x + 1)(x - 2) d y = 3(x + 1)(x + 10)(x + 5)e y = 4x2(x + 8) f y = (6x - 1)2(x + 7)

3 We20 Sketch each of the following as fully as possible.

a y = x3 + 2x2 - x - 2 b y = x3 + 6x2 + 11x + 6c y = x3 + 7x2 + 14x + 8 d y = x3 - x2 - 14x + 24e y = x3 - 3x2 - 25x - 21 f y = 3x3 + 17x2 + 28x + 12g y = 6x3 - 17x2 + 6x + 8 h y = -2x3 - 18x2

4 We21 Sketch the following using a CAS calculator. Find the intercepts and the coordinates of all turning points, correct to 2 decimal places.a y = -x3 - 8x2 - 5x + 14 b y = -x3 + 8x2 + 13x - 140c y = 3x3 - 3x2 - 15x - 9 d y = 7x3 + 29x2 + 32x + 4

exerCise

3h



5 mC Which of the following is a reasonable sketch of y = (x + 2)(x - 3)(2x + 1)?

A

2−3

y

x1–2

B

3−2

y

x1–2

−

C

−2

3

1–2

−

y

x

D

−2 3

y

x1–2

−

E

2 3

y

x1–2

6 mC The graph shown on the right could be that of:A y = x2(x + 2) B y = (x + 2)3

C y = (x - 2)(x + 2)2 D y = (x - 2)2(x + 2)E y = (x - 2)3

7 mC The graph below has the equation:

2−3 −1

−6

y

x

A y = (x + 1)(x + 2)(x + 3) B y = (x + 1)(x - 2)(x + 3)C y = (x - 1)(x + 2)(x + 3) D y = (x + 1)(x + 2)(x - 3)E y = (x - 1)(x - 2)(x - 3)

8 mC If a, b and c are positive numbers, the equation of the graph shown below is:A y = (x - a)(x - b)(x - c) B y = (x + a)(x - b)(x + c)C y = (x + a)(x + b)(x - c) D y = (x + a)(x + b)(x + c)E y = (x - a)(x + b)(x - c)

c a−b

y

x

9 mC Which of the following has only two distinct x-intercepts when graphed?A y = x(x + 1)(x + 2) B y = (x + 1)(x + 2)(x + 3)C y = x3 D y = (x + 1)(x + 2)2

E y = x(x + 1)(x - 1)

eBookpluseBookplus

Digital docSpreadsheet 014Cubic graphs

— general form

2−2

−8

y

x

136

10 a Sketch the graph of y = x3 - x2 + 3x + 5 showing all intercepts.b Factorise y = x3 - x2 + 3x + 5 expressing your answer in the form of

y = (x + a)(x2 + bx + c).c Hence, show that y = x3 - x2 + 3x + 5 has only one real solution. (Hint: Consider the

discriminant.)d If y = (x + a)(x2 + kbx + c) where k is a constant, find the values of k such that the

cubic has: i two real solutions ii three real solutions.

Note: a, b and c are the same values from part b .

Quartic graphs — intercepts methodQuartic functions are polynomial functions of degree 4.

The graphs shown below are the main types of quartic graphs.

y

x0 2

(2, 16)

y = x4 yy = (x − a)3(x − b)

repeated factor(x − a)3

xa b

y = (x − a)(x − b)(x − c)(x − d)

y

xa b c d

Negative quartics are reflected across the x-axis.Consider the general factorised quartic, f (x) = (x – a)(x – b)(x – c)

(x – d).As for the cubic functions, the Null Factor Law tells us that f (x) = 0

(that is, an x-intercept occurs) when x = a, x = b, x = c or x = d.The y-intercept occurs when x = 0, therefore the y-intercept is

f (0) = (0 – a)(0 – b)(0 – c)(0 – d) = abcd

Worked example 22

Sketch the following graphs, showing all intercepts.a y = (x - 2) (x - 1) (x + 1) (x + 3) b y = (x - 3)2 (x + 1) (x + 5) c y = (2x - 1)(x + 1)3

Think WriTe/draW

a 1 The y-intercept occurs when x = 0. a y = (x - 2)(x - 1)(x + 1)(x + 3)y = (-2)(-1)(1)(3)

= 6Point: (0, 6)

2 Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x-intercepts: if y = 0,x - 2 = 0, x - 1 = 0, x + 1 = 0 or x + 3 = 0x = 2, x = 1, x = -1 or x = -3Points: (2, 0), (1, 0), (-1, 0), (-3, 0)

3 The graph has a positive x4 coefficient, so large positive values for x result in large positive values for y.

Shape:

3i

y

xa b c d

abcd



4 Combine information from steps 1 to 3 to sketch the graph.

x

y

−3 −1 0 1 2

6

b 1 The y-intercept occurs when x = 0. b y = (x - 3)2(x + 1)(x + 5)y = (-3)2(1)(5)

= 45Point: (0, 45)


x-intercepts: if y = 0,x - 3 = 0, x + 1 = 0 or x + 5 = 0x = 3, x = -1 or x = -5Points: (3, 0), (-1, 0), (-5, 0)

3 The graph has a positive x4 coefficient. (Large positive values for x result in large positive values for y.)

Shape:

4 Combine all the information above to sketch the graph. Note that it touches the x-axis where there are repeated squared factors (x - 3)2. x

y

−5 30

45

−1

c 1 The y-intercept occurs when x = 0. c y = (2x - 1)(x + 1)3

y = (-1)(1)3

= -1Point: (0, -1)


x-intercepts: if y = 0,2x - 1 = 0 or x + 1 = 0x = 1

2 or x = -1

Points: (12, 0) and (-1, 0)

3 The graph has a positive x4 coefficient. (Large positive values for x result in large positive values for y.)

Shape:

4 Combine all the information from above to sketch the graph. Note that the graph has a point of inflection where it crosses the x-axis with repeated cubic factors (x + 1)3.

x

y

−1

−1

0 1–2

Follow the instructions in Worked example 21 to draw cubic graphs using CAS.

138

To sketch a quartic function in the form f (x) = (x – a) (x – b) (x – c) (x – d): Find the 1. y-intercept (let x = 0).Find the 2. x-intercepts (let factors of f (x) = 0).Use all available information to sketch the graph.3.

The graph shown below is y = (x – 7)(x – 3)(x + 5)(x + 8).

x

y

3 7

840

0−5−8

rememBer

Quartic graphs — intercepts method 1 We 22 Sketch the following graphs, showing all intercepts.

a y = (x – 3)(x – 2)(x + 1)(x + 2) b y = (x – 2)2(x + 1)(x + 2)c y = (x + 5)(x – 1)3 d y = (x – 1)4

e y = –x(2x – 1)(x – 3)(x + 3) f y = (x – 2)2(x + 1)2

g y = (1 – 3x)4 h y = (x + 5)3(1 – x)

2 Compare your answers to question 1 to those found using a CAS calculator.

3 If the graph of y = (x + a)(x + 3)(x + 1)(x – 3) has four distinct x-intercepts, and has a y-intercept at (0, 45), find the value of a.

4 A quartic graph has only two x-intercepts, at x = a and x = b, and a y-intercept at the point (0, 81). If a = –b:a find the equation of the quartic graph b sketch the graph, labelling all intercepts.

5 mC Which of the following has two distinct x-intercepts when graphed?A y = x(x + 7)(x – 7)(x + 2) B y = x2(x + 3)(x – 3) C y = x3(x + 27)

D y = x(x + 7)2(x – 7) E y = x4

6 mC If a, b and c are positive numbers, the equation of the graph shown is:

x−abc

y

−b ca

A y = (x + a)2(x + b)(x + c) B y = (x – a)2(x – b)(x – c) C y = (x + a)2(x – b)(x + c)D y = (x – a)(x + b)(x – c) E y = (x – a)2(x + b)(x – c)

7 mC For the graph of the quadratic equation y = (x – 2)(x + 1)(x + 3)2, the y-intercept occurs at:A 6 B –6 C –12 D –18 E 18

exerCise

3i



8 Use a CAS calculator to help you sketch the following quartics showing all intercepts and turning points correct to 2 decimal places.a y = 2x4 + x3 - 5x2 + 7x + 2 b y = 3x4 - 9x3 - 8x2 + 12x + 9c y = -x4 - 2x3 + 5x2 + 4x d y = x4 - 5x3 - 45x2 + 8x + 120e y = 3x4 - 10x2 - 3 f y = -8x4 - 10x3 + 120x2 + 15x + 35

Graphs of cubic functions in power function formRemember the power form or turning point form for quadratic graphs y = a(x - b)2 + c which was related to transformations of the basic parabola? The same understanding of transformations can be used to sketch cubic functions.

Cubic functions are also power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x, and the function is linear. When n = 2, f (x) = x2, and the function is quadratic. When n = 3, f (x) = x3, and the function is cubic. When n = 4, f (x) = x4, and the function is quartic.

Other power functions will be discussed later.Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function

is f (x) = a(x - b)n + c, (where a, b, c and n ∈ R).While all linear and quadratic polynomials are also linear and quadratic power functions, this

is not the case for cubic functions (or quartic functions). For example, a cubic power function in the form of f (x) = a(x - b)n + c has exactly one x-intercept and one stationary point of inflection. A cubic polynomial in the form f (x) = ax3 - bx2 + cx + d can have one, two or three x-intercepts and is therefore not a power function.

All cubic power functions are also cubic polynomials, but not all cubic polynomials are cubic power functions. For example, the cubic function y = 2(x - 3)3 + 1 is a polynomial and a power function. It is the graph of y = x3 under a sequence of transformations.

Such graphs have a stationary point of inflection at (b, c). A stationary point of inflection is where a graph ‘levels off’ to have a zero gradient at one point with the same sign gradient either side.

Stationary point of inflection

y y = −x3y = x3

x

y

x

Stationary point of inflection

y

x

cb

(b, c)

y = a(x - b)3 + c

Dilation factor from the x-axis (y-stretch)

x-translation y-translation

Summary of transformations

3J

140

The effect of a is illustrated below.

y

x

y = 2x3

y = x3

y = x31–2

y

x

y = −2x3

y = x3y = −x3

1–2

−

Positive a Negative a

interceptsIntercepts may be found by substituting x = 0 (to find the y-intercepts) and y = 0 (to find the x-intercepts) into the equation.

Worked example 23

Sketch the graph of each of the following, showing the stationary point of inflection and intercepts.a y = 3(x - 2)3 + 3 b y = -2x3 + 54 c y = -2(1 - 2x)3 - 16

Think WriTe/draW

a 1 Compare the equation with y = a(x - b)3 + c, having stationary point of inflection (b, c).

a y = 3(x - 2)3 + 3

2 Note the values that match, namely a = 3, b = 2 and c = 3. State the stationary point of inflection (b, c).

Stationary point of inflection (2, 3)

3 Find the y-intercept. If x = 0, y = 3(0 - 2)3 + 3 y = 3(-8) + 3 y = -21

4 Find the x-intercept. If y = 0, 0 = 3(x - 2)3 + 3 -3(x - 2)3 = 3 (x - 2)3 = -1 x - 2 = -1 x = 1

5 Note that the equation is for a positive cubic. Shape:

6 Sketch, showing the stationary point of inflection and intercepts.

y

x

−21

1 2

(2, 3)



b 1 Manipulate into y = a(x - b)3 + c form. b y = -2x3 + 54y = -2(x - 0)3 + 54

2 Note the graph is a negative cubic with stationary point of inflection (0, 54).

Stationary point of inflection (0, 54)

3 Find the y-intercept. If x = 0, y = -2(0) + 54 y = 54

4 Find the x-intercept. If y = 0, 0 = -2x3 + 54 2x3 = 54 x3 = 27 x = 3

5 Sketch, showing the stationary point of inflection and intercepts.

x

y

3

(0, 54)

c 1 Manipulate into y = a(x - b)3 + c form. c y = -2(1 - 2x)3 - 16= -2[-2(x - 1

2)]3 - 16

= -2[-8(x - 12)3] - 16

= 16(x - 12)3 - 16

2 Note the graph is a positive cubic with stationary point of inflection (1

2, -16).Stationary point of inflection (1

2, -16)

3 Find the y-intercept. If x = 0, y = -2(1 - 0)3 - 16 = -2(1) - 16 = -18

4 Find the x-intercept. If y = 0, 0 = -2(1 - 2x)3 - 16 2(1 - 2x)3 = -16 (1 - 2x)3 = -8 1 - 2x = -2 -2x = -3

x = 32

5 Sketch, showing stationary point of inflection and intercepts.

y

x

−18 ( , −16)1–2

3–2

142

Cubic power functions with y-dilation of a and stationary point of infl ection at (b, c)y = a(x - b)3 + c

y

x

(b, c)

a > 0 y

x

(b, c)

a < 0

Positive a Negative a

rememBer

Graphs of cubic functions in power function form 1 Without sketching graphs for each of the following, state:

i the dilation factor i i the coordinates of the stationary point of inflection.a y = 2(x - 1)3 + 3 b y = 3(x + 5)3 - 2c y = -2(x - 6)3 - 8 d y = -7(x + 4)3 + 1e y = (x - 9)3 + 4 f y = x3 - 7g y = -(x + 1)3 - 1 h y = 1

2(x + 2)3

i y = − 1

4 (x - 3)3 + 2 j y = 4x3

k y = − 13

x3 l y = -2x3 - 2

2 We23 Sketch the graph of each of the following, showing the stationary point of inflection and intercepts.a y = 2(x - 2)3 + 2 b y = -3(x + 3)3 + 81c y = 4(x - 4)3 - 32 d y = -5(x - 1)3 + 5e y = -x3 - 8 f y = x3 - 1g y = (x + 2)3 + 27 h y = 1

2(x + 5)3 - 32

i y = − 13

(x - 3)3 - 9 j y = − 1

4 (x + 1)3 + 2

k y = 15(x + 2)3 + 25 l = -2x3

m y = 5x3 n y = 3x3 - 3

3 Sketch the following, showing the stationary point of inflection. Intercepts are not required. Use a CAS calcu lator to verify answers.a y = (4 - x)3 + 1 b y = 3(5 - x)3 - 3c y = 2(4x - 1)3 d y = 5(3 - 2x)3 + 1

e y = 25(1 - x)3 f y = − 1

7 (3 - 4x)3 - 2

g y = -(4 - x)3 + 3 h y = (9 - 5x)3 - 7

i y = 83(6 - x)3 + 4 j y = -2(5 - 2x)3 - 1

4 mC The basic cubic graph y = x3 undergoes a dilation factor of 6 from the x-axis and is translated right 4 units and down 3 units. The equation for this graph is:A y = 6(x - 4)3 - 3 B y = 3(x - 4)3 - 6C y = -6(x - 3)3 - 4 D y = 4(x + 6)3 + 3E y = -4(x + 3)3 + 6

5 mC The graph of y = 5(2 - x)3 + 9 has a stationary point of inflection at:A (5, 2) B (5, 9) C (-2, 9)D (2, -9) E (2, 9)

exerCise

3J

eBookpluseBookplus

Digital docSpreadsheet 015Cubic graphs — basic form



6 Suggest a possible equation for each of the following, given that each is a cubic with a dilation factor of 1 or -1 from the x-axis.

a y

x

(1, 5)

b y

x

(−2, 2)

c y

x

(−3, 4)

d y

x(−3, 0)

7 Write an equation for a cubic with:a a dilation factor of 4 from the x-axis, stationary point of

inflection (2, 3)b a dilation factor of –2 from the x-axis, stationary point of

inflection (-5, 1)c a dilation factor of 1

4 from the x-axis, stationary point of

inflection (1, -2)d a dilation factor of − 1

2 from the x-axis, stationary point of

infl ection (0, 4).

domain, range, maximums and minimumsThe domain of a function is the set of x-coordinates of points on its graph. The range is the set of y-coordinates of points on the graph. Normally, the domain and range of a cubic function are the set of all real numbers, or R for short, as such graphs extend indefi nitely in both positive and negative axis directions. The domain and range of a restricted cubic function may be a smaller set of numbers.

Domain = [−5, 3]

10

−8

−5 3

Ran

ge =

[−8

, 10]

y

x

Actual maximum(within given domain)

Local maximum

eBookpluseBookplus

Digital docInvestigation

Graphs of the formy = a (x - b )n + c

3k

144

The restricted graph has a domain of x-values between -5 and 3, denoted [-5, 3]. The range is [-8, 10].

Square brackets are used to indicate that an end value is included, and correspond to a small coloured-in circle on the graph. If an end value is not included, a curved bracket is used. We show such points on a graph using a ‘hollow’ circle.

Function notationWhen we wish to convey information about the domain of a function, the following notation may be used:

f :[-4, 1] → R, where f (x) = (x - 1)(x + 2)(x + 4)

The ‘name’ of the function

The domain

The co-domain. The range is within this set.

}The rule for the function

}

Note: The range is not necessarily equal to R; the range is within R.

Worked example 24

For the function f: [-4, 1] → R where f (x) = (x - 1)(x + 2)(x + 4), sketch the graph of f (x), showing intercepts and the coordinates of any local maximum or local minimum, and state the range.

Think WriTe/display

1 Write the key information from the question.

f : [-4, 1] → R where f (x) = (x - 1)(x + 2)(x + 4)

2 Find the y-intercept. If x = 0, f (x) = (-1)(2)(4) = -8.

3 Use the Null Factor Law (or a CAS calculator) to determine x-intercepts.

If f (x) = 0, x = 1, -2 or -4.

4

Domain = [−5, 3]

10

−8

−5 3

Ran

ge =

[−8

, 10]

y

x


Use a CAS calculator to determine the local maximum and local minimum. On the Graph & Tab screen, complete the entry line as:y1 = (x - 1)(x + 2)(x + 4) | -4 ≤ x ≤ 1Tick the y1 box and tap !.Tap:• Analysis• G-Solve• Max / Min


5 Show the information in steps 1 to 4 on a sketch. Take care to draw the graph only within the required domain. The end points should be clearly visible.

y

x

(−0.21, −8.21)

(1, 0)

(−4, 0)

(−3.12, 4.06)

6 Use the graph to state the range. Range = [-8.21, 4.06].

The maximum and minimum of a graph within a certain domain are not necessarily the values of the local max imum or minimum. Sometimes an extreme value is simply the y-coordinate of an end point of a graph.

Worked example 25

Sketch f: [0, 7) → R where f (x) = x(x - 5)2, showing intercepts, end points, the local maximum and minimum, and state the range.

Think WriTe/display

1 Write the key information from the question.

f: [0, 7) → R where f (x) = x(x - 5)2

2 Find the y-intercept. If x = 0, f (x) = (0)(0 - 5)2 = 0.

3 Use the Null Factor Law (or a CAS calculator) to determine x-intercepts.

If f (x) = 0, x = 0, or x = 5.

4

Local maximumAbsolute maximum

Local minimum

Absolute minimum

y

x

Graph the function in the usual manner adjusting the viewing window settings.To determine the y-coordinate for x = 7, tap:• Analysis• G-Solve• y-CalEnter the x-value of 7 and tap:• OK

146

5 Using a CAS calculator we can establish that the absolute minimum and local minimum are at y = 0. We can also show a local maximum at x = 1 2

3.

Show the above information on a sketch.

y

x

(7, 28)

(5, 0)

(1.67, 18.52)

(0, 0)

6 Use the graph to state the range. Range = [0, 28).

Domain and range will be discussed in more detail in chapter 4, Relations and functions.A method of fi nding maximums and minimums without a CAS calculator will be covered in the study of calculus later in this book.

The absolute maximum or minimum is either the y-value at a local maximum or minimum or the y-value at an end of the domain.

rememBer

domain, range, maximums and minimumsA CAS calculator is required for this exercise.

1 State the domain and range of the sections of graph shown in each case.

a y

x

(2, –2)

(4, 0)

(–2, 5)

(–5, 2)

b y

x

(4, −2)

(1, 7)

(−2, 3)

c y

x

(2, −3)

(−4, 2)

(−3, −5)

d y

x

(4, 1)

(6, 10)

(−2, −2)

exerCise

3k

eBookpluseBookplus

Digital docSkillSHEET 3.2

Interval notation



e y

x

(5, 5)

(3, 0)(−3, 0)

(−1, 2)

f y

x

(−1, −9)

(2, 0)

(4, −5)(−3, −8)

2 We24, 25 For each of the following, sketch the graph (showing local maximums and minimums, and intercepts) and state the range.a f: [-1, 4] → R where f (x) = (x - 3)(x - 4)(x + 1)b f: [-5, 1] → R where f (x) = (x + 2)(x + 5)(x - 1)c f: [1, 3) → R where f (x) = (x - 2)2(x - 1)d f: (-3, 0] → R where f (x) = (x + 3)(x + 1)2

e f: [-8, 2) → R where f (x) = (2x - 3)(x + 1)(x + 7)f f: [0, 4] → R where f (x) = x3 - 6x2 + 9x - 4g f: [-4, -1.442] → R where f (x) = x3 + 6x2 + 11x + 6h f: (-2, 2.1) → R where f (x) = x3 + 2x2 - 5x - 6i f: [-3, 5.1] → R where f (x) = -x3 + x2 + 17x + 15j f: (-3, 21

3) → R where f (x) = 3x3 + 5x2 - 19x - 21

3 mC The range of the function shown below is:

y

x31

6

−2(2.120, −4.061)

(−0.786, 8.209)(4, 18)

A [-4.061, 18] B [-4.061, 8.209]C (0, 18] D (-2, 3]E (-2, 4]

4 mC Point A on the curve is:A an interceptB a local minimumC an absolute minimumD a local maximumE an absolute maximum

y

x

A

148

5 A roller-coaster ride is modelled by the functionf (x) = 0.001(x - 10)(x + 20)(x - 40).a What is the height above ground level of the

track at x = 50?b How far apart vertically are points A and B?

6 The course of a river as marked on a map follows the curve defined by the function f (x) = 1.5x3 - 2.7x2 + x - 1. Find the coordinates of the southernmost point on the river between x = 0 and x = 2.

−5

5

−1.5 1.5

y

x

N

modelling using technologyScientists, economists, doctors and biologists often wish to fi nd an equation that closely matches, or ‘models’ a set of data. For example, the wombat population of a particular island may vary as recorded in the following table.

Year of study (x) 0 1 2 3 4 5 6 7 8 9 10

Wombat population (W)

59 62 69 83 81 76 70 66 52 49 41

y

xA

B40

ground level−20

20

eBookpluseBookplus

Digital docInvestigation

Modelling the path of a

roller-coaster

3l eBookpluseBookplus

Interactivityint-0262

Modelling data using

polynomials



The graph below shows these data, with a possible ‘model’ for the wombat population super imposed.

y = 0.0888x3 − 2.4598x2 + 14.196x + 55.063

Year

Wom

bat p

opul

atio

n

60708090

5040302010

20 4 6 8 10

We will examine polynomial models up to degree 3, that is, models of the form:

y = a3x3 + a2x2 + a1x + a0

where a0, a1, a2 and a3 are constants. Several technological options are avail able

to assist in obtaining models for data, including a CAS calculator, spread sheets and computer algebra systems such as Mathcad. Many of these applica tions use a method involving minimising the sum of the squares of the vertical dis tances of the data points from the graph of the function — known as the least squares method.

modelling using a Cas calculatorThe CAS calculator can be used to fi nd a model or regression for a set of data. The following example employs cubic regression, but the general approach is the same for all types of regression.

Worked example 26

Fit a cubic model to the following data using a CAS calculator. Write the equation and draw a rough sketch of the graph.

x 0 1 2 3 4 5 6 7 8 9 10

y 627 545 580 528 436 318 238 229 134 169 139

Think WriTe/display

1

y = 2.6636x + 7.3182

3035

252015105

0 2 4 6 108 12

y

x

Sum of squares of theselengths is minimised.

eBookpluseBookplus

Tutorial

Worked example 26

On the Statistics screen, enter the x-values into list1 and y-values into list2. Name them ‘xcord’ and ‘ycord’ respectively.

int-0285

150

2

3

4 Write the equation. y = 1.37x3 + 19.14x2 + 8.55x + 607.48

5 Draw a rough sketch of the graph. y

x0

600

10

An equation to model data may be obtained using CAS:Enter data into the spreadsheet.1. Find the regression equation (linear, quadratic etc.).2.

rememBer

modelling using technologyUse a CAS calculator or other technology to answer the questions in this exercise.

1 We26 Find a linear model for each of the following sets of data, and draw a rough sketch of the graph.

a x 0 1 2 3 4 5 6 7 8 9 10

y -30 0 5 -9 28 29 50 68 73 77 84

exerCise

3l

eBookpluseBookplus

Digital docSpreadsheet 076

Modelling


To fi t a cubic model, tap:• Calc• Cubic RegSet: XList: main\xcord YList: main\ycord Copy Formula: y1• OK

To draw the graph of the cubic regression curve and determine its equation, enter the Graph & Tab screen and the cubic regression equation should appear as y1. To graph the equation, tap:• !


b x 0 1 2 3 4 5 6 7 8 9 10

y -15 -12 -26 -27 -12 -20 -39 -46 -50 -40 -67

c x 0 1 2 3 4 5 6 7 8 9 10

y 11 8 9 14 19 18 29 29 28 32 39

d x 0 1 2 3 4 5 6 7 8 9 10

y 53 44 39 42 35 32 30 29 23 27 19

2 Find a quadratic model for each of the following sets of data, and draw a rough sketch of the graph.

a x 0 1 2 3 4 5 6 7 8 9 10

y 19 4 48 60 36 88 126 116 159 168 122

b x 0 1 2 3 4 5 6 7 8 9 10

y 65 33 80 12 50 248 228 252 496 439 694

c x 0 1 2 3 4 5 6 7 8 9 10

y -14 16 32 36 37 51 57 56 55 54 56

d x 0 1 2 3 4 5 6 7 8 9 10

y 70 -27 92 2 -148 -327 -447 -639 -733 -910 -1204

3 Find the cubic model for each of the following sets of data, and draw a rough sketch of the graph.

a x 0 1 2 3 4 5 6 7 8 9 10

y 627 545 580 528 436 318 238 229 134 169 139

b x 0 1 2 3 4 5 6 7 8 9 10

y 21 28 91 182 81 203 345 397 730 873 1205

c x 0 1 2 3 4 5 6 7 8 9 10

y 55 84 64 29 10 4 -17 35 182 400 631

d x 0 1 2 3 4 5 6 7 8 9 10

y 45 26 109 201 399 466 621 755 800 868 854

4 For the following data set, find and sketch:a a linear model b a quadratic model c a cubic model.

x 0 1 2 3 4 5 6 7 8 9 10

y 537 681 536 624 632 763 686 885 1090 1230 1451

5 Which of the models in question 4 fits best?

6 Use the model from question 1a to predict the value of y when x = 20.

7 Use the model from question 2a to predict the value of y to the nearest unit when x = 5.5.

8 Use the model from question 3a to predict the value of y to the nearest unit when x = 12.

152

9 The value of shares in the company Mathsco is plotted by a sharemarket analyst over a 12-month period as shown.

Month0J

1J

2A

3S

4O

5N

6D

7J

8F

9M

10A

11M

Share price

0.50 0.58 0.53 0.76 1.00 1.50 1.55 2.20 3.06 3.83 4.79 4.40

a Find and sketch a quadratic model for the data.b Use your model to predict the share price 2 months later.c Give reasons why such a prediction may not be accurate.

10 The population of a colony of yellow-bellied sap-suckers on an isolated island is studied over a number of years. The population at the start of each year is shown in the table below.

Year 0 1 2 3 4 5 6 7 8 9 10

Population 250 270 310 375 410 395 335 290 290 320 325

Find and sketch a cubic model for the population, and use it to estimate the population at the start of Year 11.

Finite differencesIf pairs of data values in a set obey a polynomial equation, that equation or model may be found using the method of fi nite differences.

Consider a difference table for a general polynomial of the formy = a3x3 + a2x2 + a1x + a0.

We begin the difference table by evaluating the polynomial for x values of 0, 1, 2 etc.The differences between successive y-values (see table) are called the fi rst differences.The differences between successive fi rst differences are called second differences.The differences between successive second differences called the third differences.We will call the fi rst shaded cell (nearest the top of the table) stepped cell 1, the second

shaded cell stepped cell 2 and so on.

x y(= a3 x3 + a2 x2 + a1x + a0)

Firstdifferences

Seconddifferences

Thirddifferences

0 a0

a3 + a2 + a1

1 a3 + a2 + a1 + a0 6a3 + 2a2

7a3 + 3a2 + a1 6a3

2 8a3 + 4a2 + 2a1 + a0 12a3 + 2a2

19a3 + 5a2 + a1 6a3

3 27a3 + 9a2 + 3a1 + a0 18a3 + 2a2

37a3 + 7a2 + a1 6a3

4 64a3 + 16a2 + 4a1 + a0 24a3 + 2a2

61a3 + 9a2 + a1

5 125a3 + 25a2 + 5a1 + a0

eBookpluseBookplus

Digital docInvestigationModelling

3m



If a3 ≠ 0, the above polynomial equation represents a cubic model, and the third differences are identical (all equal to 6a3).

If a3 = 0, a2 ≠ 0 and the polynomial reduces to y = a2x2 + a1x + a0, that is, a quadratic model, and the second differences become identical (all equal to 2a2).

If a3 = 0 and a2 = 0, the polynomial becomes y = a1x + a0, that is, a linear model, and the first

differences are identical (all equal to a1).

1. Stepped cell 1 = a02. Stepped cell 2 = a1 + a2 + a33. Stepped cell 3 = 2a2 + 6a34. Stepped cell 4 = 6a3

Worked example 27

Complete a finite difference table based on the data below, and use it to determine the equation for y in terms of x.

x 0 1 2 3 4 5

y -1 0 7 20 39 64

Think WriTe

1 Place the data in columns as shown, allowing space for 3 difference columns. x y

Differences

1st 2nd 3rd

0 -1

1

1 0 6

7 0

2 7 6

13 0

3 20 6

19 0

4 39 6

25

5 64

2 Calculate the first differences and place them in the next column. The first differences are not constant, so we need to find the second differences.

3 Calculate these and place them in the next column. The second differences are constant, so our table is complete. Showing the third differences is optional. The curve is a quadratic.

4 Recall the stepped cell equations, and equate them to the shaded cells as shown:Stepped cell 1 = a0

Stepped cell 2 = a1 + a2 + a3

Stepped cell 3 = 2a2 + 6a3

Stepped cell 4 = 6a3

Here, * is used to denote solved values.

a0 = -1* [1]a1 + a2 + a3 = 1 [2] 2a2 + 6a3 = 6 [3] 6a3 = 0 [4]

5 [1] gives a0 = -1 and [4] gives a3 = 0.Substitute this information into [2] and [3].

So a3 = 0*Sub a3 = 0 into [2]:

154

a1 + a2 + 0 = 1 a1 + a2 = 1 [5]

Sub a3 = 0 into [3]:2a2 + 6 × 0 = 6

2a2 = 6 a2 = 3*

Sub a2 = 3 into [5]:

6 Substitute a2 = 0 into [5] to find a1. a1 + 3 = 1 a1 = -2*

7 Substitute your values for a0, a1, a2 and a3 into the equation: y = a3x3 + a2x2 + a1x + a0.

y = a3x3 + a2x2 + a1x + a0 becomes y = (0)x3 + (3)x2 + (-2)x + (-1) y = 3x2 - 2x - 1

The stepped equations work only if the finite differences table begins with x = 0, and x increases in steps of 1. It may be necessary on occasions to adjust the table to achieve this, as the following example shows.

Worked example 28

Complete a finite difference table based on these data and use it to determine the equation for y in terms of x.

Think WriTe

1 Construct a difference table, leaving room for the x = 0 row. x y

Differences

1st 2nd 3rd

0 -11

8

1 -3 0

8 0

2 5 0

8 0

3 13 0

8 0

4 21 0

8

5 29

2 Calculate and fill in the first differences where possible.

3 Note the first differences are constant, so last two columns are optional. The relationship is linear.

4 Working ‘backwards’, the first stepped cell must be -11 in order for the difference between it and the next cell to be 8.

5 Recall the stepped cell equations, and equate them to the shaded cells as shown:Stepped cell 1 = a0

Stepped cell 2 = a1 + a2 + a3

Stepped cell 3 = 2a2 + 6a3

Stepped cell 4 = 6a3

Here, an asterisk (*) is used to denote solved values.

a0 = -11* [1] a1 + a2 + a3 = 8 [2] 2a2 + 6a3 = 0 [3] 6a3 = 0 [4]

x 1 2 3 4 5

y -3 5 13 21 29



6 [1] gives a0 = -11, [4] yields a3 = 0, and hence [3] yields a2 = 0. Substitute this information into [2].

So a3 = 0* and a2 = 0*Sub a2 = 0 and a3 = 0 into [2]:

a1 + 0 + 0 = 8 a1 = 8* y = a3x3 + a2x2 + a1x + a0 becomes y = (0)x3 + (0)x2 + (8)x + (-11) y = 8x - 11

7 Use the asterisked values to build the equation: y = a3x3 + a2x2 + a1x + a0.

Using simultaneous equations to find a polynomial modelThe method of fitting a polynomial to a set of data using finite differences requires the data to be sequential. Often this is not the case. Simultaneous equations can be used to find a polynomial model when the data are not sequential.

The number of simultaneous equations required to find the rule of a degree n polyno mial is n + 1.

For example, to find a quadratic model, 2 + 1 = 3 points are required as a quadratic is a degree 2 polynomial. Each of the points are substituted into the general equation of the quadratic polynomial, y = ax2 + bx + c to generate 3 simultaneous equations. These can be solved using elimination or by using a CAS calculator.

Worked example 29

Using simultaneous equations, find a quadratic model for the points (-2, -9), (3, 1) and (1, 9).

Think WriTe

1 Write down the general rule of a quadratic. y = ax2 + bx + c

2 Substitute each point into the general equation to get three simultaneous equations.

a(-2)2 + b(-2) + c = -9∴ 4a - 2b + c = -9 [1]

a(3)2 + b(3) + c = 1∴ 9a + 3b + c = 1 [2]

a(1)2 + b(1) + c = 9∴ a + b + c = 9 [3]

3 Solve equations [1], [2] and [3] using elimination.Equation [1] - [2] to eliminate c.Equation [2] - [3] to eliminate c.Equation [4] × 8Equation [5] × 5Add equations [6] and [7] to eliminate a and solve for b.Substitute b = 4 into equation [4] to find a.

Substitute a = -2 and b = 4 into equation [3] to find c.

-5a - 5b = -10 [4] 8a + 2b = -8 [5] -40a - 40b = -80 [6] 40a + 10b = -40 [7] -30b = -120

∴ b = 4 -5a - 5(4) = -10 -5a = 10

∴ a = -2 -2 + 4 + c = 9

∴ c = 7

4 Write the rule. y = -2x2 + 4x + 7

156

Worked example 30

Using simultaneous equations, find a cubic model for the points (-2, -10), (1, 2), (3, -20) and (6, 22). Use a CAS calculator to solve the simultaneous equations.

Think WriTe/display

1 Write down the general rule of a cubic. y = ax3 + bx2 + cx + d

2 Substitute each point into the general equation to get four simultaneous equations.

a(-2)3 + b(-2)2 + c(-2) + d = -10∴ -8a + 4b - 2c + d = -10 [1]

a(1)3 + b(1)2 + c(1) + d = 2∴ a + b + c + d = 2 [2]

a(3)3 + b(3)2 + c(3) + d = -20∴ 27a + 9b + 3c + d = -20 [3]

a(6)3 + b(6)2 + c(6) + d = 22∴ 216a + 36b + 6c + d = 22 [4]

3

The values are a = 1, b = -5, c = -4 and d = 10.

4 Write the rule. y = x3 - 5x2 - 4x + 10

5


To solve the simultaneous equations using a CAS calculator, on the Main screen and using the soft keyboard, tap:• )• {NComplete the entry lines as:-8a + 4b - 2c + d = -1027a + 9b + 3c + d = 227a + 9b + 3c + d = -20216a + 36b + 6c + d = 22Type in the variables to be solved in the bottom right corner, a, b, c, d, and then press E.

An alternative way to solve simultaneous equations with a CAS calculator is the use the Define command. Tap:• Action• Command• DefineComplete the entry line as:Define f (x) = ax3 + bx2 + cx + d.Press E, and then tap:• )• {NComplete the entry lines as shown on the right.Type in the variables to be solved in the bottom right corner, a, b, c, d, and then press E.


To determine the polynomial equation using the method of finite differences:1. (a) set up a table and find the first, second and third differences(b) equate the stepped cell equations to the appropriate cells in the table(c) find the coefficients a0, a1, a2, a3 for the equation y = a3x3 + a2x2 + a1x + a0.Stepped cell 1: a0 Stepped cell 2: a1+ a2

+ a3

Stepped cell 3: 2a2+ 6a3 Stepped cell 4: 6a3

Alternatively, simultaneous equations can be used to find a polynomial model. 2. Simultaneous equations can be solved with or without a CAS calculator.

rememBer

Finite differences 1 We27, 28 For each of the following, complete a finite difference table based on the data

below, and use it to determine the equation of y in terms of x.

a x 0 1 2 3 4 5

y 6 17 28 39 50 61

b x 0 1 2 3 4 5

y 100 74 48 22 -4 -30

c x 0 1 2 3 4 5

y -4 7 32 71 124 191

d x 1 2 3 4 5

y 1 -9 -13 -11 -3

e x 0 1 2 3 4 5

y -7 -10 -1 26 77 158

f x 0 1 2 3 4 5

y 16 17 20 31 56 101

g x 0 1 2 3 4 5

y -23 -11 5 25 49 77

h x 1 2 3 4 5

y 28 27 32 31 12

i x 0 1 2 3 4 5

y -27 -18 -9 0 9 18

j x 1 2 3 4 5

y -7 -3 -3 -7 -15

exerCise

3m

158

k x 0 1 2 3 4 5

y -66 -5 58 99 94 19

l x 0 1 2 3 4 5

y 43 35 27 19 11 3

2 Triangular numbers may be illustrated as shown at right.

If x is the number of dots on the base of each diagram, and y is the total number of dots:a complete the table belowb find an equation linking x and y

x (base dots) 0 1 2 3 4 5

y (total dots)

c find the total number of cans in the supermarket display shown at right using the equation found in b , and check your answer by counting the cans.

3 The diagonals in polygons of various types are shown below in

red. Find the relationship between the number of dots (x) and the number of diagonals (n).(Hint: Continue patterns in a difference table so that it is completed back to x = 0.)

4 If n is the number of different squares that can be found within a square grid of edge

length x, find an equation for n in terms of x, and use this equation to find the number of different squares on a chessboard.

x = 1n = 1

x = 2n = 5

x = 3n = ?

5 Find a linear model for the following sets of points.a (2, 1) (-1, -23) b (-4, 6) (8, -3)

6 We29 Using simultaneous equations, find a quadratic model for the following sets of points.a (-2, -13) (6, -37) (-4, -57) b (-1, 4) (1, -2) (4, 19)c (4, 8) (0, 8) (-4, 24) d (-5, -360) (-2, -96) (6, -272)

7 We30 Using simultaneous equations, find a cubic model for the following sets of points. Use a CAS calculator to solve the simultaneous equations.a (-6, 3) (-3, -27) (3, -33) (2, 3) b (-2, -39) (1, 6) (4, 141) (-3, -118)c (4, -10) (6, 90) (8, 302) (-2, 2) d (-1, -4) (1, -8) (4, -314) (0, -6)

8 Using simultaneous equations, find a quartic model for the following sets of points. Use a CAS calculator to solve the simultaneous equations.a (1, 2) (-3, 354) (4, 1313) (2, 79) (-1, -2)b (-4, 73) (0, 1) (2, -11) (-2, 13) (6, -707)

x = 4

x = 2x = 1

x = 3

158 maths Quest 11 mathematical methods Cas for the Casio Classpad


sUmmary

Expandi ng

When expanding three linear factors:• expand two factors first, then multiply the result by the remaining linear factor1. collect like terms at each stage2. (3. x + 2)3 may be written as (x + 2)(x + 2)(x + 2).

Long division of polynomials

Long division of polynomials is similar to long division with numbers.• The highest power term is the main one considered at each stage.• Key steps are:•

How many?1. Multiply and write the result underneath.2. Subtract.3. Bring down the next term.4. Repeat until no pronumerals remain to be divided.5. State the quotient and the remainder.6.

Polynomial values

P• (a) means the value of P(x) when x is replaced by a and the polynomial is evaluated.

The remainder and factor theorems

Remainder • R = P(a), when P(x) is divided by x - a.If • P(a) = 0, then (x - a) is a factor of P(x).

Factorising polynomials

To factorise a polynomial:• let 1. P(x) = the given polynomialuse the factor theorem to find a linear factor2. use long or short division to find another factor3. repeat steps 2 and 3, or factorise by inspection if possible.4.

Alternatively, use the factor function on your CAS calculator.

Sum and difference of two cubes

a• 3 + b3 = (a + b)(a2 - ab + b2)a• 3 - b3 = (a - b)(a2 + ab + b2)

Solving polynomial equations

To solve a polynomial equation:• rewrite the equation so it equals zero1. factorise the polynomial as much as possible2. let each linear factor equal zero and solve for 3. x in each case.

Cubic graphs — intercepts method

To sketch a cubic function of the form • f (x) = Ax3 + Bx2 + Cx + D:determine if the expression is a positive or negative cubic (that is, if 1. A is positive or negative)find the 2. y-intercept (let x = 0)factorise if necessary or possible, for example, obtain an expression in the form 3. f (x) = (x - a)(x - b)(x - c)

160

4. find the x-intercepts (let factors of f (x) equal 0)use all available information to sketch the graph.5.

Quartic graphs — intercepts method

• To sketch a quartic function in the form f (x) = (x – a) (x – b)(x – c) (x – d):

find the 1. y-intercept (f (0) = abcd)find the 2. x-intercepts (let factors of f (x) = 0)use all available information to sketch the graph.3.

Graphs of cubic functions in power function form

Cubic with • y-dilation of a and stationary point of inflection at (b, c) y = a(x - b)3 + c

y

x

(b, c)

a > 0

Positive a

y

x

(b, c)

a < 0

Negative a

Domain, range, maximums and minimums

The absolute maximum or minimum is either the • y-value at a local maximum or minimum, or the y-value at an end of the domain.

Modelling using technology

An equation to model data may be obtained using a CAS calculator.• Enter data as lists.1. Find the regression equation (linear, quadratic etc.).2.

20

4−1 5–2

y

x

x

y

3 7

840

0–5–8



Finite differences

To use the method of finite differences.• Set up a table as shown, and find differences by subtracting successive values 1. (value – previous value). Circle or shade the stepped cells.

x yFirst

differencesSecond

differencesThird

differences

0 Stepped cell 1

Stepped cell 2

1 Stepped cell 3

Stepped cell 4

2

3 Etc.

Etc.

4 Previous value

Value –previous value

5 Value

Use the following equations to determine the polynomial model’s coefficients:2. 1. Stepped cell 1 = a0

Stepped cell 2 2. = a1 + a2 + a3

Stepped cell 3 3. = 2a2 + 6a3

Stepped cell 4 4. = 6a3

Simultaneous equations

Simultaneous equations can be used to find a polynomial model, given a number of points.• The number of simultaneous equations required to find the rule of a degree • n polynomial is n + 1.Simultaneous equations can be solved with or without a CAS calculator.•

} Equation of the polynomial model is y = a3x3 + a2x2+ a1x + a0

162

ChapTer reVieW

shorT ansWer

1 Expand: a (x - 2)2(x + 10) b (x + 6)(x - 1)(x + 5) c (x - 7)3 d (5 - 2x)(1 + x)(x + 2).

2 Find the quotient and remainder when the first polynomial is divided by the second in each case. a x3 + 2x2 - 16x - 3, x + 2 b x3 + 3x2 - 13x - 7, x - 3 c -x3 + x2 + 4x - 7, x + 1

3 If P(x) = -3x3 + 2x2 + x - 4, find: a P(1) b P(-4) c P(2a).

4 Without dividing, find the remainder when x3 + 3x2 - 16x + 5 is divided by x - 1.

5 Show that x + 3 is a factor of x3 - 2x2 - 29x - 42.

6 Factorise x3 + 4x2 - 100x - 400.

7 Factorise: a 1 - 125x3 b (x - 2)3 + (x + 3)3.

8 Solve: a 5(x + 5)3 + 5 = 0 b (2x + 1)2 (x - 3)2 = 0 c x3 - 9x2 + 26x - 24 = 0.

9 Sketch: a y = x(x - 2)(x + 11) b y = x3 + 6x2 - 15x + 8 c y = -2x3 + x2.

10 Sketch: a y = x(x – 7) (x – 2) (x + 4) b y = (2x – 1) (x + 1) (x + 4)2

c y = -x(x + 5)3.

11 Sketch y = − 1

8 (x + 1)3 + 8.

12 Find the range of f :[-6, 3] → R, where f (x) = (x + 1)(2 - x)(x + 5).

13 Complete a finite difference table, and use it to determine the equation for y in terms of x for the following data set.

x 0 1 2 3 4 5

y 8 7 8 17 40 83

14 The following series of diagrams show the maximum number of regions produced by drawing chords in a circle.

x = 0r = 1

x = 1r = 2

x = 2r = 4

x = 3r = 7

Find a relationship between the number of chords (x) and the maximum number of regions (r).

15 Using simultaneous equations, find a cubic model for the points (-1, -10), (2, -4), (-3, -104) and (0, -2).

16 Use the remainder theorem to determine if 2x2 - 3x3 + 7x + 11 is exactly divisible by (x + 1).

[VCAA 2003]

mUlTiple ChoiCe

1 The expansion of (x + 5)(x + 1)(x - 6) is:A x3 - 30B x3 - 6x2 + 5x - 6C x3 + 12x2 - 31x + 30D x3 - 31x - 30E x3 + 5x2 - 36x - 30

2 x3 + 5x2 + 3x - 9 is the expansion of:A (x + 3)3

B x(x + 3)(x - 3)C (x - 1)(x + 3)2

D (x - 1)(x + 1)(x + 3)E (x + 1)(x + 2)(x - 3)

Questions 4 and 5 refer to the following long division.

x2 + x + 2x + 2xx + 4))x3 + 5x2 + 6x - 1

x3 + 4x2

x2 + 6xx2 + 4x

2x 2x 2 - 12x 2x 2 + 8

-9

exam Tip A surprisingly large number of students did not know the remainder theorem. Many used long division and received no marks as they had not responded to the explicit instruction to use the remainder theorem. Others used both without indicating which was the solution to the question asked. P(1) appeared on occasions. Some students found P(-1) and then made no statement as to what this meant.

[VCAA Assessment report 1 2003]



3 The quotient is:A -9 B 4 C x + 4D x2 + x + 2 E x3 + 5x2 + 6x - 1

4 The remainder is:A -9 B 2 C 4D 2x - 1 E 2x + 8

5 If P(x) = x3 - 3x2 + 7x + 1, then P(-2) equals:A -34 B -33 C -9D 7 E 35

6 The remainder when x3 - 7x is divided by x - 1 is:A -6 B 1 C 6D 7 E 8

7 Which of the following is a factor of x3 - 3x2 - 18x + 40?A (x - 4) B (x - 2) C (x + 1)D (x + 3) E (2x - 1)

8 x3 + 6x2 - 15x + 8 factorises to:A (x - 1)2(x + 8) B (x + 1)2(x + 8)C (x + 2)3 D (x + 1)(x + 2)(x + 4)E (x - 1)(x + 2)(x + 4)

9 64x3 - y3 factorises to:A (4x - y)(16x2 + 4xy + y2)B (4x - y)(16x2 - 4xy + y2)C (4x - y)(16x2 + 8xy + y2)D (4x + y)(16x2 - 8xy + y2)E (4x + y)(16x2 - 4xy + y2)

10 Which of the following is the solution to -(x - 4)3 - 2 = 6?A -6 B -2 C 2 D 4 E 6

11 Which of the following is a solution to (x - 11)(3x + 5)(7 - 3x)(2x + 5) = 0?A -11 B − 3

5 C 37

D 53 E 7

3

12 The equation for this graph could be:A y = (x - 5)(x + 1)(x + 3)B y = (x - 3)(x - 1)(x + 5)C y = (x - 3)(x + 1)(x + 5)D y = (3x - 1)(x + 1)(x - 5)E y = (5 - x)(1 + x)(3 + x)

5−3 −1

y

x

13 The equation for the graph shown below could be: A y = (x – 3)2 (x + 3)2

B y = (x – 3) (x + 3)3

C y = (x – 3)3 (x + 3)D y = (x – 3)2 (x + 3)2

E y = (x + 3)4

30 x

y

–3

–81

14 Which of the following shows the graph of y = -2(x + 5)3 - 12?

A

(−5, −12)

y

x

B y

x

(−5, 12)

C

(5, −12)

y

x

D (5, 12)

y

x

E y

x(−5, −12)

164

Questions 15 and 16 refer to the following graph (below).

−5 2 7

y

x

(4.813, 60.370)

(−2.147, −108.222)

(−7, 252)

15 The domain of the graph is:A [-108.222, 252) B (-2.147, 4.183]C (-108.222, 60.370] D (-7, 7]E (-6, 7]

16 The range of the graph is:A [-108.222, 252) B [-2.147, 4.183]C [-108.222, 60.370] D [7, 252)E [0, 252)

17 The data below obey which type of relationship?

x 0 1 2 3 4 5

y 0 4 16 66 208 520

A Linear B Quadratic C Cubic D Quartic E None of the above

18 Which of the following points lies on the curve of the quadratic model that fits the points (1, 0), (0, -7) and (2, 11)?A (3, -4) B (0, 7) C (-2, -7) D (-1, -10) E (-1, -14)

19 A polynomial function p has degree three. A part of its graph, near the point on the graph with coordinates (2, 0), is shown below.

2.22.121.91.8 x

Which one of the following could be the rule for the third degree polynomial p?A p(x) = x(x + 2)2

B p(x) = x(x - 2)3

C p(x) = x2(x - 2)D p(x) = (x - 1)(x - 2)2

E p(x) = -x(x - 2)2

[VCAA 2003]

20 Let f be a polynomial function of degree 3. The graph of the curve with rule y = f (x) either intersects or touches the x-axis at exactly two points (a, 0) and (b, 0). A possible rule for f could be:A f (x) = (x - a)(x - b)B f (x) = (x - a)(x + b)2

C f (x) = (x - a)(x - b)2

D f (x) = (x + a)2(x - b)E f (x) = (x + a)2(x + b)

[VCAA 2004]

21 Which one of the following is a complete set of linear factors of the third-degree polynomial ax bxax bx3 3− − , where a and b are positive real numbers?A x, ax2 - bB x, ax - b , ax + b

C x, ax b−D x, a x - b, a x + b

E x, a x - b, a x + b

[VCAA 2004]

22 Part of the graph y = ax3 + bx2 + cx + d is shown below.

y

x0

1020304050

−20−10

−30−40

1 2 3 4 5−2−3−4 −1

The values of a, b and c respectively could be:

a b c d

A 1 -2 -11 12

B 2 -4 -22 24

C 1 2 -22 12

D -2 -6 -2 24

E 3 -1 1 24

[VCAA 2006]



exTended response

1 For P(x) = 5x3 - 3x2 - 6x - 22, find P(3) and P(-x).

2 Find the value of m if x + 3 is a factor of 2x3 - 15x2 + mx - 21.

3 Factorise x3 - 2x2 - 9x + 18. Sketch the graph of f (x) = x3 - 2x2 - 9x +18.

4 Factorise (3x - 2)3 + (x + 5)3.

5 Determine the x- and y-intercepts of the cubic graph y = (2 - 3x)(4x + 1)(2x - 7). Hence, sketch the graph.

6 The graph y = x3 has been moved parallel to the x-axis 5 units to the left and moved upwards 2 units from the x-axis.What is the equation of the translated graph and what are the coordinates of the point of inflection?Sketch the translated graph.

7 Sketch the graph of y = (x + 2)2(x - 3)(x - 4), showing all intercepts.

8 The polynomial P(x) = x3 + ax2 + bx + 54 is exactly divisible by x - 9 and also exactly divisible by x - 6.a Find the values of a and b.b Find the third factor.c Hence, sketch the graph of the polynomial y = x3 + ax2 + bx + 54.

9 Factorise x3 - 2x2 - 3x + 6 over the real number field. Sketch the graph of y = x3 - 2x2 - 3x + 6.

10 Find the points of intersection between y = x3 - x2 - 19x - 13 and 3x + y - 7 = 0.

11 Use the method of finite differences to fit a polynomial model to the following data.

x 0 1 2 3 4

y 4 16 25 30 30

12 A diagram of a proposed waterslide based on a cubic function appears below. Find:a the height, h1, of the top of the slideb the coordinates of point A (where the slide enters the water)c the length, L, of the ladderd the height, h2, of the ‘mini-hump’ to the nearest centimetre.

y

x(–5, 0) (8, 0.22)

y = −0.008(x3 − 30x2 + 285x − 900)

A

h2

h1L

13 An innovative local council decides to put a map of the district on a web site. Part of the map involves two key features — the Cubic River and the Linear Highway. A mathematically able web site designer has found the following equations for these features: Cubic River: y = x3 + x2 - 4x - 4 Linear Highway: y = 5x + 5.a Sketch the river and highway, showing x- and y-axis intercepts.b Find the coordinates of the points of intersection of the highway and the river.c A fun-run organiser wishes to arrange checkpoints at the closest points of intersection. Find the distance

between the proposed checkpoints.

166

14 A cubic function in the form f (x) = ax3 + bx2 + cx + d has the following values.

x 0 1 2 3 4 5

y 42 36 20 0 -18 -28

a Use fi nite differences to fi nd the values of a, b, c and d.b State one factor of f (x), giving your reasoning.c Using long or short division, factorise f (x).d Sketch the graph of f (x), labelling all intercepts.

15 The height (in centimetres) of a wave above a 1-metre pole is measured over an interval of 8 seconds. The wave’s height has been found to approximate the function H = t3 – 13t2 + 48t.a Find the initial height.b Using a CAS calculator, sketch the function and fi nd the local maximum and minimum height of

the wave.c The height of a later wave is found to approximate the function

K = t3 – 14t2 + 53t – 40. Show at what times the height of this wave is exactly the same height as the pole.

eBookpluseBookplus

Digital docTest YourselfChapter 3



eBookpluseBookplus aCTiViTies

Chapter openerDigital doc

10 Quick Questions: Warm up with ten quick • questions on cubic and quartic functions (page 109)

3B Long division of polynomialsTutorial

We 4 • int-0281: Watch how to perform long division of polynomials (page 113)

3C Polynomial valuesDigital docs

Spreadsheet 016: Investigate solutions to cubic • equations (page 117)Work• SHEET 3.1: Review the discriminant (page 117)

3D The remainder and factor theoremsDigital doc

Spreadsheet 016: Investigate solutions to cubic • equations (page 120)

3E Factorising polynomialsDigital docs

SkillSHEET 3.1: Practise using calculating and • using the discriminant (page 121)Spreadsheet 096: Investigate zeros of cubics • (page 124)

Tutorial

We 12 • int-0282: Use long division to factorise a cubic (page 121)

3F Sum and difference of two cubesTutorial

We 15 • int-0283: Watch how to factorise expressions using the sum or difference of two cubes formulae (page 125)

3G Solving polynomial equationsDigital doc

Work• SHEET 3.2: Factorising cubics and quartics using long division, applying the Null Factor Law to determine x-intercepts and sketching cubics and quartics (page 129)

3H Cubic graphs — intercepts methodTutorial

We 20 • int-0284: Sketch the graph of a cubic showing axial intercepts (page 132)

Digital doc

Spreadsheet 014: Investigate the effect of changing • coeffi cients of cubics in general form on its graph (page 135)

3J Graphs of cubic functions in powerfunction form

Digital docs

Spreadsheet 015: Investigate the graphs of cubic • functions in power form (page 142)Investigation: Graphs of the form • y = a(x - b)n + c (page 143)

3K Domain, range, maximums and minimumsDigital docs

SkillSHEET 3.2: Practice expressing intervals using • varying notation (page 146)Investigation: Modelling the path of a roller-coaster • (page 148)

3L Modelling using technologyInteractivity int-0262

Modelling data using polynomials: Use the • interactivity to consolidate your understanding of how to fi t a polynomial model to data (page 148)

Tutorial

We 26 • int-0286: Watch how to fi t a cubic model to a set of data using a CAS calculator (page 149)

Digital docs

Spreadsheet 076: Investigate the best model for a set • of data (page 150)Investigation: Fitting a model exactly • (page 152)

Chapter reviewDigital doc

Test Yourself: Take the end-of-chapter test to test • your progress (page 166)

To access eBookPLUS activities, log on to

www.jacplus.com.au

int-0285

168

exam praCTiCe 1 ChapTers 1 To 3

shorT ansWer 25 minutes

1 Determine the exact values for x for which x2 - 5 = 3x. 3 marks

2 Let f:[-3, 1] → R, where f (x) = 2x3 + 5x2 - 4x - 3. a Show that (x - 1) is a linear factor. 2 marksb When factorised, f (x) = (x - 1)(x + 3)(ax + b).

Determine the values of a and b. 3 marksc Hence, sketch the graph of f (x). 3 marks

3 The line y = ax + b passes through the point (2, 1) and is parallel to the line y - 4x + 3 = 0. Determine the values of a and b. 2 marks

4 Let f : [-1, 3] → R, f (x) = 2x2 - 4x - 3.a Express f in the form f (x) = a(x - b)2 + c.

Hence, state the coordinates of the turning point. 3 marks

b State the range for f. 2 marksc Determine the exact values of the x-intercepts.

3 marks d Sketch f on a set of axes. Label all key features.

3 marks

mUlTiple ChoiCe 12 minutes

1 A straight line passes through the points (2, 4) and (-1, -5). Its equation would be:A y = 6 - x B y = x + 2 C 3y = x + 10D y = 3x - 2 E y = -3x + 10

2 The exact values of x for which 3(2x - 1)2 + 2(2x - 1) - 8 = 0 are:A (2x + 1)(6x - 7)B (3x + 4)(x - 2)

C − 43

or 2

D 1310

or −1310

E −12

or 76

3 If (x - a) is a linear factor of the function f (x) = x3 + 2x2 - 11x - 12, then the possible value for a is:A -3 B -1 C 0D 1 E 4

4 A cubic function has the following axis intercepts: x-intercepts 1, 3 and -4 and y-intercept 24. The equation that would best describe this function would be: A 24(x + 1)(x + 3)(x - 4) B (x + 1)(x + 3)(x - 4) + 24C (x - 1)(x - 3)(x + 4) + 24D 2(x - 1)(x - 3)(x + 4)E 2 (x + 1)(x + 3)(x - 4)

5 Three linear functions are defined as: L1: 4y - 3x - 8 = 0L2: 3y + 4x + 1 = 0L3: 3y - 4x - 2 = 0

Which one of the following statements is correct?A L1 and L2 are parallel. B L1 and L3 are parallel.C L1 and L2 are perpendicular.D L1 and L3 are perpendicular.E L1, L2 and L3 are parallel.

6

b−a

y

x

Which one of the following rules best describes the graph above?A y = (x + a)2 (x - b) B y = (x - a)2 (x - b)C y = (x + a) (x - b) D y = (x - b)2 (x + a)E y = (x - a)2 (x - b)

7 A polynomial of degree three passes through the origin and has x-intercepts at -2 and 3. The equation for this polynomial would best be described by which one of the following?A y = (x - 2)(x + 3) B y = (x + 2)(x - 3)C y = x(x - 2)(x + 3) D y = x(x + 2)(x - 3)E y = (x - 1)(x - 2)(x + 3)

8 A linear function L(x) = ax + 1 intersects the quadratic function Q(x) = 2x2 - x - 1 at the point (-0.5, 0). The coordinates of the second point of intersection would be closest to:A (-1, 1) B (-1, 2) C (-1, 4)D (2, 3) E (2, 5)


169exam practice 1

exTended response 40 minutes

1 Points A (-2, 1), B (1, 5) and C (5, 2) are vertices on a triangle. a Show that the triangle is an isosceles triangle. 3 marks b Determine the equation of the perpendicular bisector of the line AC. 4 marksc Show that the perpendicular bisector found in part b passes through vertex B. 2 marks

2 Seng and Victor are training for a 1000-m race. The distance, in metres, they each ran was recorded in 1 minute intervals. Victor was delayed for 1 minute because he was tying up his shoelaces when he was 110 m from the starting line. The individual distances are recorded in the table below.

Time (minutes) 0 1 2 3 4 5

Seng’s distance (m) 0 200 380 680 880 1040

Victor’s distance (m) -110 0 200 560 875 1050

a The relationship between distance and time for Victor can be modelled using a quartic function. Determine this quartic function, Qv. Write your function in terms of Qv (distance in metres) and t (minutes). Express coeffi cients correct to 2 decimal places. 2 marks

b Seng’s distance and time relationship can be modelled using a cubic function. Determine this cubic function, Cs. Write your function in terms of Cs (distance in metres) and t (minutes). Express coeffi cients correct to 2 decimal places. 2 marks

c i State the domain of Qv. 1 mark ii State the domain of Cs. 1 mark iii Sketch both Qv and Cs on the same set of axes. 3 marksd Using your functions Qv & Cs, determine the time fi rst Victor passes Seng. Write your answer in

minutes and seconds. 2 marks

3 A children’s playground in the shape of a rectangle is to be constructed at a local park. The longest length is 5 metres longer than the shorter side length.a If x is the shortest side, write down an expression for the longest side in terms of x. 1 markb Write down an equation for the area, A(x), of the playground in terms

of x. 2 marksc If the maximum area of the playground is 150 m2, determine the

dimensions of the playground. 3 marksA miniature bike path will be built within the playground. It will travel from the gate to the seesaws, slides and swings. The bike path can be modelled by the cubic function: B(x) = x3 - 6x2 +10x, where x is the horizontal distance, in metres, and B(x) is the vertical distance, in metres, from the gate.d Assuming that the slides are on the path, determine the vertical distance

the slides are from the gate, if their measurement is 3 metres horizontally from the gate. 2 marks

e If the seesaws are on the bike path, show that the seesaws are located 1 metre horizontally and 5 metres vertically from the gate. 2 marks

f Determine the shortest distance, in metres, between the seesaws and the slides. Write your answer in exact form. 2 marks x

y

0

eBookpluseBookplus

Digital docSolutions

Exam practice 1

3A 3B 3C 3D 3E 3F Cubic and 3G 3H 3I quartic 3J 3K ...mathsbooks.net/JACPlus Books/11 Methods/Ch03...

Documents

Transcript of 3A 3B 3C 3D 3E 3F Cubic and 3G 3H 3I quartic 3J 3K ...mathsbooks.net/JACPlus Books/11 Methods/Ch03...